UCNRLF
EH MEMOEIAM
Edward Bright
(Brcenlcafs Jttathematical 3ttk^
UNIVERSITY ALGEBRA.
DESIGNED FOR THE USE OF SCHOOLS
AND COLLEGES.
PREPARED BY
WEBSTER WELLS, S. B.,
i
ASSISTANT PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS
INSTITUTE OF TECHNOLOGY;
LEACH, SHEWELL, AND SANBORN,
BOSTON AND NEW YORK.
COPYRIGHT, 1880.
q/v
By WEBSTER WELL?..
PREFACE.
This work was designed to take the place of Green
leaf's Higher Algebra, portions of which have been used
in the preparation of the present volume. It contains
the topics usually taught in High Schools and Colleges,
and the author's aim has been to present the subject in a
compact form and in clear and concise language. The
principles have been developed with regard to logical ac
curacy, and care has been given to the selection of exam
ples and practical illustrations which should exercise the
student in all the common applications of the algebraic
analysis. The full treatment given in the earlier chap
ters renders the previous study of a more elementary
textbook unnecessary.
Attention is invited to the following chapters, including
those in which the most important changes have been
made in the Higher Algebra : —
Parentheses.
Factoring.
Zero and Infinity.
Theory of Exponents.
Simultaneous Equations involving Quadratics.
Binomial Theorem for Positive Integral Exponents.
Undetermined Coefficients.
Logarithms.
The answers have been put by themselves in the back
part of the book, and those have been omitted which, if
79795
i / PREFACE.
given, would destroy the utility of the problem. The ex
amples are over eighteen hundred in number, and are pro
gressive, commencing with simple applications of the
rules, and passing gradually to those which require some
thought for their solution.
The works of Todhunter and Hamblin Smith, and other
standard volumes, have been consulted in the preparation
of the work, and have furnished a number of examples
and problems. The author has also received numerous
suggestions from practical teachers, to whom he would
here express his thanks.
WEBSTER WELLS.
Boston, 1884.
UNIVERSITY ALGEBRA.
CONTENTS.
Chapter Paob
I. Definitions and Notation 1
Symbols of Quantity 1
Symbols of Operation 2
Symbols of Relation 4
Symbols of Abbreviation 5
Algebraic Expressions 5
Axioms 8
Negative Quantities 12
II. Addition 14
III. Subtraction 19
IV. Use of Parentheses 21
V. Multiplication 24
VI. Division 31
VII. Formula 38
VIII. Factoring 40
IX. Greatest Common Divisor 53
X. Least Common Multiple 61
XI. Fractions 66
Reduction of Fractions 70
Addition and Subtraction of Fractions 78
Multiplication of Fractions 82
Division of Fractions 85
Complex Fractions 87
XII. Simple Equations. — One unknown quantity 89
Transformation of Equations 91
Solution of Equations 95
Vlll
CONTENTS.
XIII. Problems. — One unknown quantity 103
XIV. Simple Equations. — Two unknown quantities 113
Elimination 115
XV. Simple Equations. — More than two unknown quan
tities 123
XVI. Problems. —More than one unknown quantity 127
Generalization of Problems 133
XVII. Discussion of Problems 137
Interpretation of Negative Results 139
XVIII. Zero and Infinity 142
Problem of the Couriers 143
XIX. Inequalities 148
XX. Involution 153
Involution of Monomials 153
Involution of Polynomials 154
Square of a Polynomial 155
Cube of a Binomial 156
Cube of a Polynomial 157
XXI. Evolution 158
Evolution of Monomials 159
Square Root of Polynomials 160
Square Root of Numbers 163
Cube Root of Polynomials 168
Cube Root of Numbers 171
Any Root of Polynomials 174
XXII. The Theory of Exponents 176
XXIII. Radicals 188
Reduction of Radicals 1 88
Addition and Subtraction of Radicals 193
Multiplication of Radicals 194
Division of Radicals 196
Involution of Radicals 197
Evolution of Radicals 198
Reduction of Fractions with Irrational Denominators 199
Imaginary Quantities 202
Quadratic Surds 205
Radical Equations 208
CONTENTS. ix
XXIV. Quadratic Equations. — One unknown quantity... 210
Pure Quadratic Equations 211
Affected Quadratic Equations 213
XXV. Problems. — Quadratic Equations. — One unknown
quantity 223
XXVI. Equations in the Quadratic Form 227
XXVII. Simultaneous Equations involving Quadratics 233
XXVIII. Problems. — Quadratic Equations. — Two unknown
QUANTITIES 244
XXIX. Theory of Quadratic Equations 249
Discussion of the General Equation 249
XXX. Discussion of Problems leading to Quadratic
Equations 257
Interpretation of I maginary Results 259
Problem of the Lights 259
XXXI. Ratio and Proportion 262
XXXII. Variation 270
XXXIII. Arithmetical Progression 274
XXXIV. Geometrical Progression 282
XXXV. Harmonical Progression 291
XXXVI. Permutations and Combinations 294
XXXVII. Binomial Theorem. — Positive Integral Exponent 298
XXXVIII. Undetermined Coefficients 304
Expansion of Fractions into Series 307
Expansion of Radicals into Series 310
Decomposition of Rational Fractions 312
Reversion of Series 318
XXXIX. Binomial Theorem. —Any Exponent 321
XL. Summation of Infinite Series 328
Recurring Series 328
The Differential Method 332
Interpolation 336
x CONTENTS.
XLI. Logarithms 339
Properties of Logarithms 342
Use of the Table 348
Solutions of Arithmetical Problems by Logarithms 354
Exponential Equations 358
Application of Logarithms to Problems in Compound In
terest 359
Exponential and Logarithmic Series 362
Arithmetical Complement 366
XLII. General Theory of Equations 369
Divisibility of Equations 370
Number of Roots . 371
Formation of Equations 373
Composition of Coefficients 374
Fractional Roots 376
Imaginary Roots 376
Transformation of Equations 377
Descartes' Rule of Signs 383
Derived Polynomials 385
Equal Roots 386
Limits of the Roots of an Equation 389
Sturm's Theorem 392
XLIII. Solution of Higher Numerical Equations 399
Commensurable Roots 400
Recurring or Reciprocal Equations 404
Cardan's Method for the Solution of Cubic Equations. .... 408
Biquadratic Equations 411
Incommensurable Roots 412
Horner's Method 412
Approximation by Double Position 417
Newton's Method of Approximation 419
Answers to Examples 421
Table of the Logarithms of Numbers from 1 to 10,000. .Appendix.
ALGEBRA.
I. — DEFINITIONS AND NOTATION.
1. Quantity is anything that can be measured ; as dis
tance, time, weight, and number.
2. The Measurement of quantity is accomplished by find
ing the number of times it contains another quantity of the
same kind, assumed as a standard. This standard is called
the unit of measure.
3. Mathematics is the science of quantities and their re
lations.
4. Algebra is that branch of mathematics in which the
relations of quantities are investigated, and the reasoning
abridged and generalized, by means of symbols.
5. The Symbols employed in Algebra are of four kinds:
symbols of quantity, symbols of operation, symbols of relation,
and symbols of abbreviation.
SYMBOLS OF QUANTITY.
6. The Symbols of Quantity generally used are the
figures of Arithmetic and the letters of the alphabet.
The figures are used to represent known quantities and
determined values, and the letters any quantities whatever,
known or unknown.
7. Known Quantities, or those whose values are given,
■2 ALGEBRA.
when not expressed by figures, are usually represented by the
first letters of the alphabet, as a, b, c.
8. Unknown Quantities, or those whose values are not
given, are usually represented by the last letters of the
alphabet, as x, y, z.
9. Zero, or the absence of quantity, is represented by the
symbol 0.
10. Quantities occupying similar relations in different op
erations are often represented by the same letter, distinguished
by different accents, as a', a", a'", read "a prime," "a second,"
"a third," etc. ; or by different subscript figures, as a 1} a 2) a s ,
read " a one/' " a two," " a three," etc.
SYMBOLS OF OPERATION.
11. The Symbols of Operation are certain signs or char
acters used to indicate algebraic operations.
12. The Sign of Addition, + , is called "plus." Thus,
a ( b, read " a plus b" indicates that the quantity b is to be
added to the quantity a.
13. The Sign of Subtraction, — , is called "minus."
Thus, a — b, read " a minus b" indicates that the quantity
b is to be subtracted from the quantity a.
The sign ~ indicates the difference of two quantities when
it is not known which of them is the greater. Thus, a ~ b
indicates the difference of the two quantities a and b.
14. The Sign of Multiplication, x , is read " times"
"into," or "multiplied by." Thus, aXb indicates that the
quantity a is multiplied by the quantity b.
A simple point (. ) is sometimes used in place of the sign X
The sign of multiplication is, however, usually omitted, except
between two arithmetical figures separated by no other sign;
multiplication is therefore indicated by the absence of any
sign. Thus, 2ab indicates the same as 2 X a X b, or 2 . a . b.
DEFINITIONS AND NOTATION. 3
15. The quantities multiplied are called factors, and the
result of the multiplication is called the product.
16. The Sign of Division, ~, is read "divided by."
Thus, a 4 b indicates that the quantity a is divided by the
quantity b.
Division is otherwise often indicated by writing the divi
dend above, and the divisor below, a horizontal line. Thus,
 indicates the same as a i b. Also, the sign of division
b
may be replaced in an operation by a straight or curved line.
Thus, a I b, or b ) a, indicates the same as a f b.
17. The Exponential Sign is a figure or letter written at
the right of and above a quantity, to indicate the number of
times the quantity is taken as a factor. Thus, in x 3 , the 3 in
dicates that x is taken three times as a factor ; that is, x 3 is
equivalent to xxx.
The product obtained by taking a factor two or more times
is called a power. A single letter is also often called the
first power of that letter. Thus,
a 2 is read "a to the second power," or "a square," and
indicates a a;
a 3 is read " a to the third power," or " a cube," and indi
cates acta;
a 4 is read " a to the fourth power," or " a fourth," and indi
cates a a a a;
a n is read " a to the nth power," or " a nth," and indicates
a a a etc., to n factors.
The figures or letters used to indicate powers are called
exponents ; and when no exponent is written, the first power
is understood. Thus, a is equivalent to a 1 .
The root of a quantity is one of its equal factors. Thus,
the root of a 2 , a 3 , or a* is a.
18. The Radical Sign, \f , when prefixed to a quantity,
indicates that some root of the quantity is to be extracted.
4 ALGEBRA.
Thus,
SJ a indicates the second or square root of a;
^]a indicates the third or cuhe root of a;
$ a indicates the fourth root of a; and so on.
The index of the root is the figure or letter written over
the radical sign. Thus, 2 is the index of the square root, 3 of
the cube root ; and so on.
When the radical sign has no index written over it, the
index 2 is understood. Thus, sj a is the same as tf a.
SYMBOLS OF RELATION.
19. The Symbols of Relation are signs used to indicate
the relative magnitudes of quantities.
20. The Sign of Equality, =, read " equals" or "equal
to," indicates that the quantities between which it is placed
are equal. Thus, x = y indicates that the quantity x is equal
to the quantity y.
A statement that two quantities are equal is called an
equation. Thus, x \ 4=2 x — 1 is an equation, and is read
" x plus 4 equals 2x minus 1."
21. The Sign of Ratio, : , read " to," indicates that the
two quantities between which it is placed are taken as the
terms of a ratio. Thus, a : b indicates the ratio of the quan
tity a to the quantity b, and is read " the ratio of a to b."
A proportion, or an equality of ratios, is expressed by writ
ing the sign =, or the sign : :, between equal ratios. Thus,
30 : 6 = 25 : 5 indicates that the ratio of 30 to G is equal to
the ratio of 25 to 5, and is read "30 is to 6 as 25 is to 5."
22. The Sign of Inequality, > or < , read " is greater
than" or "is less than" respectively, when placed between
two quantities, indicates that the quantity toward which the
opening of the sign turns is the greater. Tims, x > y is
read "x is greater than y" ; x—6< y is read "x minus 6
is less than y."
DEFINITIONS AND NOTATION. 5
23. The Sign of Variation, cc, read "varies as," indicates
that the two quantities between which it is placed increase
or diminish together, in the same ratio. Thus, a oc _ is read
" a varies as c divided by d."
SYMBOLS OF ABBREVIATION.
24. The Signs of Deduction, .. and v , stand the one for
therefore or hence, the other for since or because.
25. The Signs of Aggregation, the vinculum , the
bar  , the parenthesis ( ) , the brackets [ ] , and the braces j £,
indicate that the quantities connected or enclosed by them are
to be subjected to the same operations. Thus,
a + b X oc, a x, (a + b) x, [a + fr]x, \a + bc x,
b
all indicate that the quantity a + b is to be multiplied
by x.
26. The Sign of Continuation, , stands for and so
on, or continued by the some law. Thus,
a, a + b, a + 2 b, a + 3 b, is read
"a, a plus b, a plus 2b, a plus 3 b, and so on."
ALGEBRAIC EXPRESSIONS.
27. An Algebraic Expression is any combination of alge
braic symbols.
28. A Coefficient of a quantity is a figure or letter pre
fixed to it, to show how many times the quantity is to be
taken. Thus, in 4«, 4 is the coefficient of a, and indicates
that a is taken four times, or a + a + a + a. Where any
number of quantities are multiplied together, the product of
6 ALGEBEA.
any of them may be regarded as the coefficient of the product
of the others; thus, in abed, ab is the coefficient of ed,
b of ac d, a b d of c, and so on.
When no coefficient of a quantity is written, 1 is understood
to he the coefficient. Thus, a is the same as 1 a, and x y is
the same as 1 x y.
29. The Terms of an algebraic expression are its parts
connected by the signs + or — . Thus,
a and b are the terms of the expression a + b ;
2 a, b 2 , and — 2 a c, of the expression 2 a + b 2 — 2 a c.
30. The Degree .of a term is the number of literal factors
which it contains. Thus,
2 a is of the first degree, as it contains but one literal factor,
a & is of the second degree, as it contains two literal factors.
3 a b 2 is of the third degree, as it contains three literal factors.
The degree of any term is determined by adding the expon
ents of its several letters. Thus, a b 2 c s is of the sixth degree.
31. Positive Terms are those preceded by a plus sign ; as,
+ 2 a, or + a b 2 .
When a term has no sign written, it is understood to be
positive. Thus, a is the same as + a.
Negative Terms are those preceded by a minus sign ; as,
— 3 a, or —be.
This sign can never be omitted.
32. In a positive term, the coefficient indicates how many
times the quantity is taken additively (Art. 28) ; in a nega
tive term, the coefficient indicates how many times the quan
tity is taken svbtractively. Thus,
+ 2 x is the same as + x + x ;
— 3 a is the same as — a — a — a.
DEFINITIONS AND NOTATION. 7
33. If the same quantity be both added to and subtracted
from another, the value of the latter will not be changed ;
hence if any quantity b be added to any other quantity a, and
b be subtracted from the result, the remainder will be a;
that is,
# a + b — b = a.
Consequently, equal terms affected by unlike signs, in an
expression, neutralize each other, or cancel.
34. Similar or Like Terms are those which differ only in
their numerical coefficients. Thus,
2 x y 2 and — 1 xif are similar terms.
Dissimilar or Unlike Terms are those which are not similar.
Thus,
b x 2 y and bxy 2 are dissimilar terms.
35. A Monomial is an algebraic expression consisting of
only one term ; as, 5 a, 7 a b, or 3 b' 2 c.
A monomial is sometimes called a simple quantity.
36. A Polynomial is an algebraic expression consisting of
more than one term ; as, a + b, or 3 a 2 + b — 5 b 3 .
A polynomial is sometimes called a compound quantity, or a
multinomial.
37. A Binomial is a polynomial of two terms ; as,
a — b, 2 a + b 2 , ov dad 2 — b.
A binomial whose second term is negative, as a — b, is some
times called a residual.
38. A Trinomial is a polynomial of three terms ; as,
a + b + c, or a b + c 2 — b 3 .
39. Homogeneous Terms are those of the same degree ; as,
a 2 , 3 be, and — 4 x 2 .
$ ALGEBRA.
40. A polynomial is homogeneous when all its terms are
homogeneous ; as, a 3 + 2 a b c — 3 b 3 .
41. A polynomial is said to be arranged according to the
decreasing powers of any letter, when the term having the
highest exponent of that letter is placed first, that having
the next lower immediately after, and so on. Thus,
a 3 +3a 2 b + 3ab 2 + b 3
is arranged according t6 the decreasing powers of a.
A polynomial is said to be arranged according to the increas
ing powers of any letter, when the term having the lowest
exponent of that letter is placed first, that having the next
higher immediately after, and so on. Thus,
a 3 + 3 a 2 b + 3 a b 2 + b 3
is arranged according to the increasing powers of b.
42. The Reciprocal of a quantity is 1 divided by that
quantity. Thus, the reciprocal of
a is  , and of x + y is
a x + y
43. The Interpretation of an algebraic expression consists
in rendering it into an arithmetical quantity, by means of the
numerical values assigned to its letters. The result is called
the numerical value of the expression.
Thus, the numerical value of
4d+ 3 be — d
when a = 4, b = 3, c = 5, and d = 2, is
4x4 + 3x3x52 = 16 + 45  2 = 59.
AXIOMS.
44. An Axiom is a selfevident truth.
Algebraic operations are based upon definitions, and the
following axioms : —
DEFINITIONS AND NOTATION. 9
1. If the same quantity, or equal quantities, be added to
equal quantities, the sums will he equal.
• 2. If the same quantity, or equal quantities, he subtracted
from equal quantities, the remainders will he equal.
3. If equal quantities be multiplied by the same quantity, or
by equal quantities, the products will be equal.
4. If equal quantities be divided by the same quantity, or
by equal quantities, the quotients will be equal.
5. If the same quantity be both added to and subtracted from
another, the value of the latter will not be changed.
6. If a quantity be both multiplied and divided by another,
the value of the former will not be changed.
7. Quantities which are equal to the same quantity are equal
to each other.
8. Like powers and like roots of equal quantities are equal.
9. The whole of a quantity is equal to the sum of all its
parts.
EXERCISES ON THE PRECEDING DEFINITIONS AND
PRINCIPLES.
45. Translate the following algebraic expressions into
ordinary language :
d
m
1. 3 a 2 + b c — q. 5. cd : — = ab : \J x*.
3
n
x
2. 4 m . 6. (a — b)x = [c + d~] y.
< } 3a — d
" 2c + b
3. ^ a + b = ^ a 2 — c. 7. {m + r — s}n =
4,mn>pa. 8. \J ^ < (e  d) (h + fj.
46. Put into the form of algebraic expressions the follow
ing :
1. Five times a, added to two times b.
2. Two times x, minus y to the second power.
10 ALGEBRA.
3. The difference of x and y.
4. The product of «, b, c square, and d cube.
5. x + y multiplied by a — b.
6. a square divided by the sum of b and c.
7. x divided by 3, increased by 2, equals three times y,
diminished by 11.
8. The reciprocal of a + b, plus the square of a } minus the
cube root of b, is equal to the square root of c.
9. The ratio of 5 a divided by b, to d divided by c square,
equals the ratio of x square y cube to y square z fourth.
10. The product of in and a + b is less than the reciprocal
of x cube.
11. The product of x + y and x — y is greater than the
product of the square of a — d into the cube of a + b.
12. The quotient of a divided by 3 a — 2 is equal to the
square root of the quotient of m + n divided by 2x — y 2 .
47. Find the numerical values of the following : —
When a = 6, b = 5, c = 4, and d — 1, of
1. a 2 + 2 a b — c + d. 4. a 2 (a + b) — 2abc.
2. 2a?2a 2 b + c 3 . 5. 5a 2 b±ab 2 + 21c.
3. 2a 2 + 3bc5. 6. 7 a 2 + (a b) (ac).
When a = 4, b = 2, c = 3, and d = l, of
a
2 J,2
1.15a7(b+c)d. 10. ^ +  + ^.
8. 25a 2 7(b 2 + c 2 ) + d\ 11. 4 + 1 .
„« £» c '25 a — 30 c— d
9.  H 1  . 12. .
bed b + c
When a = \, b = \, c = \, and x = 2, of
IS. (2 a + 3 b + 5 c) ($ a + 3 b  5 e) (2 a 3 b + 15 c).
DEFINITIONS AND NOTATION. H
15. x*  (2 a + 3 b) x 3 + (3 a  2 b) x 2  ex + be.
When a=b, and b = £, of
16 5 a + ^  C 3 a  ( 2 a  ^)]
17 13 a + 3 b + {7 Q + b) + [3 a + 8 (4 a  b)~\)
2a + Sb
When b = 3, c = 4, d = 6, and e = 2, of
18. V 27T v' 27+ y/2Z 19. V 3^7+ ^"9^^27.
When « = 16, & = 10 ; cc = 5, and y=l, of
20. (b  x) (y/a + b) + ^ (a  b) (x + y).
48. What is the coefficient of
1. x in 3 n 2 x ? 3. x y in — 20 m? xyz z ?
2. a c s in a J 2 c 3 d 4 ? 4. ra 2 w 3 in 5 a 8 m 2 a; ?i 3 ?
What is the degree of
5. 3ax? 6. 2m*nx*? 7. a 2 b s c 2 d 5 ? 8. 2mcr 2 y 3 .??
Arrange the following expressions according to the increas
ing powers of x :
9. 2cc 2 3a; + x 3 + l4a; 4 .
10. 3 x y* — 5 x s y + y i — x 4 — x 2 y 2 .
Arrange the following expressions according ' o the decreas
ing powers of a :
11. 1 a 2 2 a + a 3 + 2 a*.
12. aJ 3 i 4 + a 4 4« 2 i 2 3a 3 5.
12 ALGEBRA.
NEGATIVE QUANTITIES.
49. The signs + and — , besides indicating the operations
of addition and subtraction, are also used, in Algebra, to indi
cate the nature or quality of the quantities to which they are
prefixed.
To illustrate, let us suppose a person, having a property of
$ 500, to lose $ 150, then gain $ 250, and finally to incur a debt
of $ 450 ; it is required to find the amount of his property.
Since gains have an additive effect on property, and debts or
losses a subtractive effect, we may indicate these different
qualities algebraically by prefixing the signs + and — to them,
respectively ; thus, we should represent the transactions as
follows,
$ 500  $ 150 + $ 250  $ 450 ;
which reduces to $ 150, the amount required.
But suppose, having a property of $ 500, he incurs a debt
of $ 700 ; we should represent the transaction algebraically as
follows,
$5001700;
or, as incurring a debt of $ 700 is equivalent to incurring two
debts, one of $ 500 and the other of $ 200, the transaction may
be expressed thus,
$ 500  $ 500  $ 200.
Now since, by Art. 33, $ 500 and — $ 500 neutralize each
other, we have remaining the isolated negative quantity
— $200 as the algebraic representative of the required prop
erty. In Arithmetic, we should say that he owed or was in
debt $200; in Algebra, we make also the equivalent state
ment that his property amounts to —$200.
In this way we can conceive the possibility of the indepen
dent existence of negative quantities; and as, in Arithmetic,
losses may be added, subtracted, multiplied, etc., precisely as
though they were gains, so, in Algebra, negative quantities
DEFINITIONS AND NOTATION. 13
may be added, subtracted, multiplied, etc., precisely as though
they were positive.
The distinction of positive and negative quantities is applied
in a great many cases in the language of everyday life and in
the mathematical sciences. Thus, in the thermometer, we
speak of a temperature above zero as +, and one below as — ;
for instance, +25° means 25° above zero, and —10° means
10° below zero. In navigation, north latitude is considered
j, and south latitude — ; longitude west of Greenwich is con
sidered +, and longitude east of Greenwich — ; for example,
a place in latitude — 30°, longitude + 95°, would be in latitude
30° south of the equator, and in longitude 95° west of Green
wich. And, in general, when we have to consider quantities
the exact reverse of each other in quality or condition, we
may regard quantities of either quality or condition as posi
tive, and those of the opposite quality or condition as negative.
It is immaterial which quality we regard as positive ; but hav
ing assumed at the commencement of an investigation a certain
quality as positive, we must retain the same notation through
out.
The absolute value of a quantity is the number represented
by that quantity, taken independently of the sign affecting it.
Thus, 2 and — 2 have the same absolute value.
But as we consider a person who owns $ 2 as better off
than one who owes $2, so, in Algebra, we consider + 2 as
greater than — 2 ; and, in general, any positive quantity,
however small, is considered greater titan any negative quan
tity.
Also, as we consider a person who owes $ 2 as better off than
one who owes $3, so, in Algebra, we consider — 2 as greater
than —3; and, in general, oftivo negative quantities, that is
regarded as the greater which has the less number of units, or
which has the smaller absolute value.
Again, as we consider a person who has no property or debt
as better off than one who is in debt, so, in Algebra, zero is
considered greater than any negative quantity.
14 ALGEBRA.
II. — ADDITION.
50. Addition, in Algebra, is the process of collecting two
or more quantities into one equivalent expression, called, the
sum.
51. In Arithmetic, when a person incurs a debt of a certain
amount, we regard his property as diminished by the amount
of the debt. So, in Algebra, using the interpretation of nega
tive quantities as given in Art. 49, adding a negative quantity
is equivalent to subtracting an equal positive quantity. Thus,
the sum of a and — b is obtained by subtracting b from a, giv
ing as a result a — b.
Hence, the addition of monomials is indicated by uniting
the quantities with their respective signs. Thus, the sum of
a, — b, c, d, — e, and — f, is
a—b+c+d—e —f.
The addition of polynomials is indicated by enclosing them
in parentheses (Art. 25), and uniting the results with + signs.
Thus, the sum of a + b and c — d is
(a + b) + (c — d).
52. Let it be required to add c — d to a + b.
If we add c to a + b, the sum will be a + b + c. But we
have to add to a + b a quantity which is d less than c. Conse
quently our result is d too large. Hence the required sum will
be a + b + c diminished by d, or a + b + c — d.
Hence, the addition of polynomials may also be indicated by
uniting their terms with their respective signs.
53. Let it be required to add 2 a and 3 a.
By Art. 32, 2 a = a + a,
and 3 a = a + a + a.
ADDITION. 15
Hence (Art. 52) the sum of 2 a and 3 a is indicated by
a \ a + a + a + a,
which, by Art. 32, is equal to 5 a. Hence, 2a + 3a = 5a.
54. Let it be required to add — 3 a and — 2 a.
By Art. 32, — 3 a = — a — a — a,
and — 2 a = — a — a.
Hence (Art. 52), the sum of — 3 a and — 2 a is indicated by
— a — a — a — a — a,
or — 5 a (Art. 32). Hence, — 3 a — 2 a = — 5 a.
From our ideas of negative quantities (Art. 49), we may ex
plain this result arithmetically as follows :
If a person has two debts, one of $ 3 and the other of $ 2,
he may be considered to be in debt to the amount of $ 5.
55. Let it be required to add 4 a and — 2 a.
4 a = a + a+ a + a,
and — 2« = — a — a.
Hence, the sum of 4 a and — 2 a is indicated by
a + a + a + a — a — a.
Now, by Art. 33, the third and fourth terms are neutralized
by the fifth and sixth, leaving as the result a + a, or 2 a.
Hence, 4 « — 2 a = 2 a.
We may explain this result arithmetically as follows :
If a person has $4 in money, and incurs a debt of $2, his
property may be considered to amount to $ 2.
56. Let it be required to add — 4 a and 2 a.
— 4:a = — a — a — a — a,
and 2a = ffl+«.
16 ALGEBRA.
Hence, the sum of — 4 a and 2 a is indicated by
— a — a — a — a + a + a.
The third and fourth terms neutralize the fifth and sixth,
leaving as the result — a — a or —2a. Hence,
— 4ta + 2a = — 2a.
We may explain this result arithmetically as follows :
If a person has $ 2 in money, and incurs a debt of 84, he
may be considered to be in debt to the amount of $2.
.57. From Arts. 55 and 56 we derive the following rule
for the addition of two similar (Art. 34) terms of opposite
sign:
To add two similar terms, the one positive and the other
negative, subtract the smaller coefficient from the larger, affix
to the result the common symbols, and prefix the sign of the
larger.
For example, the sum of 7 x y and — 3xy is 4:xy,
the sum of 3 a 2 b s and — 11 a~ b 3 is — 8 a 2 b 3 .
58. In Arithmetic, when adding several quantities, it
makes no difference in which order we add them ; thus,
3 + 5 + 9, 5 + 3 + 9, 9+3 + 5, etc., all give the same result,
17. So also in Algebra, it is immaterial in what order the
terms are united, provided each has its proper sign. Thus,
— b + a is the same as a — b.
Hence, in adding together any number of similar terms,
some positive and some negative, we may add the positive
terms first, and then the negative, and finally combine these
two results by the rule of Art. 57.
Thus, in finding the sum of 2 a, — a, la, 6 a, — 4 a, and
— 5 a, the sum of the positive terms 2 a, 7 a, and 6 a, is 15 a,
and the sum of the negative terms — a, — 4 a, and — 5 a,
is — 10 a ; and the sum of 15 a and — 10 a is 5 a.
59. Let it be required to add 6 a — 7 x, 3 x — 2 a + 3 y,
and 2 x — a — mn.
ADDITION. 1 7
We might obtain the sum in accordance with Art. 52, by
uniting the terms by their respective signs, and combining
similar terms by the methods previously given. It is however
customary in practice, and more convenient, to set the expres
sions down one underneath the other, similar terms being in
the same vertical column ; thus,
6 a — 1 x
— 2 a + 3 x + 3 y
— a + 2 x —mn
3 a — 2 x + 3 y — m n.
It should be remembered that only similar terms can be
combined by addition ; and that the algebraic sum of dissimilar
terms can only be indicated by uniting them by their respective
signs.
60. From the preceding principles and illustrations is de
rived the following
RULE.
To add together two or more expressions, set them down one
underneath the other, similar terms being in the same vertical
column. Find the sum of the similar terms, and to the result
obtained unite the dissimilar terms, if any, by their respective
signs.
EXAMPLES.
1.
2.
3.
4.
5.
la
— 6 m
13 n
— 4 a x
2a 2 b
3 a
m
n
— 3 ax
ab
a
— 11 m
— 20n
a x
11 a b
5 a
— 5m
6 n
— lax
5a 2 b
11a
— m
8n
— ax
±a 2 b
a
20 m
— n
12 ax
9aH
18 ALGEBRA.
6. 7. 8.
la — mp 2 2 a — 3 x ab + c d
a + 6 mp 2 — a + 4x — ab \ cd
— 11 a — 3 mp 2 a + x 3 a b — 2 cd
8 a + 11 m j9 2 5 a — 7 a; lab — 5 cd
— 9 a— 1 mp 2 —4: a— x —4tab + 6cd
18 a — 15 mp 2 —3 a + 1 x 2 ab — 5cd
Find the sum of the following :
9. 4,xy z, — 3xy z, — 5xy z, 6x y z, — 9xy z, and 3 x y z.
10. 5 m n 2 — 8x 2 y, — m n 2 + x 2 y, — 6m n 2 — 3x 2 y, 4:mn 2
+ 1 x 2 y, 2 m n 2 + 3 x 2 y, and — ra ri 2 — 2x 2 y.
11. 3a 2 + 2ab + 4,b 2 , 5a 2 Sab + b 2 , a 2 +5abb 2 ,
18 a 2 20 ab 19 b 2 , and 14 a 2  3 a b + 20 b 2 .
12. 2a — 5b — c +1, 3b — 26a + 8c, c + 3a4, and
1 + 2 b  5 c.
13. 6x — 3y+lm, 2 n — x + y, 2 y — 4x— 5 m, and
m + n — y.
14. 2 a  3 b + 4 <7, 2 &  3 d + 4 c, 2 cZ  3 e + 4 a, and
2c3a + 4i.
15. 3 cc — 2 y — z, 3 y — 5 x — 1 z, 8 z — y — x, and 4 x.
16. 2 m — 3n + 5r — t, 2 n — 6 t — 3 r — m, r + 3 m — 5n
+ 4t, and 3 t — 2 r + 1 n — 4 m.
17. 4:inn + 3 ab — 4 c, 3 x — 4 a b + 2 m n, and 3 m 2 — 4 p.
18. 3 a + b — 10, c — d — a, —4c + 2a — 3b — l, and
4 a; 2 + 5  18 m.
19. 4a;8_5 a 8_ 5ax 2 +6a 2 x, 6a s + 3x* + Aax 2 + 2a 2 x,
11 X s +19 ax 2  15 a 2 x, and 10 x 3 + 1 a 2 x + 5 a 3  18 a x 2 .
20. la — 5if, S^x + 2a, oif — \/x, and — 9a + 7tfx.
21. 3 a b + 3 (« + b),  a b + 2 (a + b), 7 a b — 4 (a + b),
and — 2 a b + 6 (a + b).
22. lsjy4(ab), 6 \J y + 2 (a  b), 2 ^ y + (ab), and
sjy — 3(a — b).
SUBTRACTION. 19
III. — SUBTRACTION.
61. Subtraction, in Algebra, is the process of finding one
of two quantities, when their sum and the other quantity are
given.
Hence, Subtraction is the converse of Addition.
The Minuend is the sum of the quantities.
The Subtrahend is the given quantity.
The Remainder is the required quantity.
As the remainder is the difference between the minuend
and subtrahend, subtraction may also be defined as the process
of finding the difference between two quantities.
62. Subtraction may be indicated by writing the subtra
hend after the minuend, with a — sign between them. Thus,
the subtraction of b from a is indicated by
a — b.
In indicating subtraction in this way, the subtrahend, if a
negative quantity or a polynomial, should be enclosed in a
parenthesis. Thus, the subtraction of —b from a is indi
cated by
and the subtraction of b — c from a by
a— (b — c).
63. Let it be required to subtract b — c from a.
According to the definition of Art. 61, we are to find a
quantity which when added to b — c will produce a ; this
quantity is evidently a — b + c, which is the remainder re
quired.
Now, if we had changed the sign of each term of the sub
trahend, giving — b + c, and had added the resulting expres
sion to a, we should have arrived at the same result, a — b + c.
20 ALGEBRA.
Hence, to subtract one quantity from another, we may change
the sign of each term of the subtrahend, and add the residt to
the minuend.
64. 1. Let it be required to subtract 3 a from 8 a.
According to Art. 63, the result may be obtained by adding
— 3 a to 8 a, giving 5 a (Art. 55).
2. Subtract 8 a from 3 a.
By Art. 63, the result is 3 a — S a or —5a (Art. 56).
3. Subtract — 2 a from 3 a.
Result, 3 a + 2 a or 5 a.
4. Subtract 3 a from —2 a.
Result, — 2a3«or —5 a.
5. Subtract — 2 a from —5 a.
Result, — 5 a + 2 a or —3 a.
6. Subtract —5a from —2 a.
Result, — 2 a + 5 a or 3 a.
65. In Arithmetic, addition always implies augmentation,
and subtraction diminution. In Algebra this is not always
the case ; for example, in adding — 2 a to 5 a the sum is 3 a,
which is smaller than 5 a ; also, in subtracting —2 a from 5 a
the remainder is 7 a, which is larger than 5 a. Thus, the
terms Addition, Subtraction, Sum, and Remainder have a
much more general signification in Algebra than in Arith
metic.
66. From Art. 63 we derive the following
RULE.
To subtract one expression from another, set the subtrahend
underneath the minuend, similar terms being in the same ver
tical column. Change the sign of each term of the subtrahend
from + to — , or from — to + , and add the restdt to the
minuend.
USE OF PAKENTHESES. 21
EXAMPLES.
1.
2.
3.
4.
5.
27 a
17 x
13 3,
 10 m n
5 a 2 b
13 a
11b
Ay 
 18 m n
UaH
ab + cd — ax 7x+5y—3a
Aab — 3 cd\ Aax x—7y+ha—A
8. 9.
7 abcllx + 5y 48 5 \Ja 3 y 2 + 7 a 6
llabc+ 3x + 7?/ + 100 3<Ja+ y 2 5a7
10. Subtract —5b from  12 J.
11. From 31 x 2  3 y 2 + a b take 17 x 2 + 5 if  4 a b + 7.
12. Subtract a — b + c from a + b — c.
13. Subtract 6a — 3^ — 5c from Qa\3b — 5 c + 1.
14. From 3m5ft+r2s take 2 r + 3 n — m — 5 s.
15. Take 4 a — b + 2 i — 5 d from a" — 3 & + a  c.
16. From m 2 + 3 n z take — 4 m' 2 — 6 ?i 8 + 71 cc.
17. From a + b take 2«25 and — a + b.
18. From a — b — c take — a + b + c and a — b + c.
IV. — USE OF PARENTHESES.
67. The use of parentheses is very frequent in Algebra,
and it is necessary to have rules for their removal or introduc
tion
22 ALGEBRA.
68. Let it be required to indicate the addition of 3 a. and
5 b — c + 2 d ; this we may do by placing the latter expression
in a parenthesis, prefixing a + sign, and writing after the
former quantity, thus :
3 a + (5 b  c + 2 d).
If the operation be performed, we obtain (Art. 60),
3a + 5b — c + 2d.
69. Again, let it be required to indicate the subtraction of
5b — c\2d from 3 a ; this we may do by placing the former
expression in a parenthesis, prefixing a — sign, and writing
after the latter quantity, thus :
3a(5bc + 2d).
If the operation be performed, we obtain (Art. 66),
3 a — 5 b + c — 2d.
70. It will be observed that in the former case the signs
»f the terms within the parenthesis are unchanged when the
parenthesis is removed ; while in the latter case the sign of
each term within is changed, from + to — , or from — to +.
Hence, we have the following rule for the removal of a paren
thesis :
If the parenthesis is preceded by a f sign, it may be re
moved if the sign of every enclosed term be unchanged; and
if the parenthesis is preceded by a — sign, it may be removed
if the sign of every enclosed term be changed.
71. To enclose any number of terms in a parenthesis, we
take the reverse of the preceding rule :
Any number of terms may be enclosed in a parenthesis, with
a + sign prefixed, if the sign of every term enclosed be un
changed ; and in a parenthesis, with a — sig?i prefixed, if
the sign of every term enclosed be changed.
USE OF PARENTHESES. 23
72. As the bracket, brace, and vinculum (Art. 25) have the
same signification as the parenthesis, the rules for their re
moval or introduction are the same. It should be observed
in the case of the vinculum, that the sign apparently prefixed
to the first term underneath is in reality the sign of the vin
culum ; thus, + a — b signifies + (a — V), and — a — b signi
fies — (a — b).
73. Parentheses will often be found enclosing others ; in
this case they may be removed successively, by the preceding
rule ; and it is better to begin by removing the inside pair.
74. 1. Remove the parentheses from 3 a — (2 a — 5) —
(a + 7).
Result, 3a — 2a + 5 + a — 7 — 2a — 2.
2. Remove the parentheses etc., from
6 a  [3 a + (2 a  { 5 a  [4 a  a  2] } )].
In accordance with Art. 73, we remove the vinculum first,
and the others in succession. Thus,
6 a — [3 a +(2 a {5 a [4 a a 2]})]
= 6 a  [3 a + (2 a  {5 a  [4 a  a + 2]})]
= 6 a — [3 a + (2 a— {5 a — 4 a + a — 2})]
= 6 a — [3 a + (2 a — 5 a + 4 a — a + 2)]
= Qa — [3a + 2a — 5a + 4a — a + 2^\
= 6a — 3a — 2a + 5a — 4:a + a — 2 = 3a — 2, Ans.
3. Enclose the last three terms of a — b~ c + d+ e —f in
a parenthesis with a — sign prefixed.
Result, a — b — c—(—d — e+f).
24 ALGEBRA.
EXAMPLES.
Remove the parentheses, etc., from the following :
4. a — (b — c) + (d — e).
5. 3a(2a{a + 2}).
6. 5 x — (2 x — 3 y) — (2 x +. 4 y).
7. a — b + c — {a + b — c) — {c + b — a).
8. in' — 2n+ (a — n + 3 ?m 2 ) — (5 a + 3 w — m 2 ).
9. 2 m  [n — {3 wi (2w m) } ].
10. 8x — (5x — [4 a; — ?/ — &■]) — (— a; — 3 y).
11. 2«[5J+ {3c(a+[2ft3a + 4c])}].
12. 3c+(2a[5c{3a + c4a}]).
13. 6 a  [5 a  (4 a  {  3 a  [2 a a 1]})].
14. 2 m  [3 m  (5 m  2)  { m  (2 wi 3m + 4)}].
75. As another application of the rule of Art. 70, we have
the following four results :
+ (+ a) is equivalent to + a ;
+ ( — a) is equivalent to — a ;
— (+ a) is equivalent to — a ;
— (—a) is equivalent to + a.
V. — MULTIPLICATION.
76. Multiplication, in Algebra, is the process of taking
one quantity as many times as there are units in another
quantity.
The Multiplicand is the quantity to be multiplied or taken.
The Multiplier is the quantity by which we multiply.
The Product is the result of the operation.
The multiplicand and multiplier are often called factors.
MULTIPLICATION. 25
77. The product of the factors is the same, in whatever
order they are taken.
For we know, from Arithmetic, that the product of two
numbers is the same, in whatever order they are taken ; thus
we have 3 X 4 or 4 X 3 eacli equal to 12. Similarly, in Alge
bra, where the symbols represent numbers, we have a X b or
b X a each equal to a b (Art. 14).
78. Let it be required to multiply a — b by c.
By Art. 77, multiplying a — b by c is the same as multiply
ing c by a — b. To multiply c by a — &, we multiply it first
by a, and then by b, and subtract the second result from the
first, e multiplied by a gives a c, and multiplied by b gives
b c. Subtracting the second result from the first we have
a c — b c
the product required.
79. Let it be required to multiply a — b by c — d.
To multiply a — b by c — d, we multiply it first by c, and
then by d, and subtract the second result from the first. By
Art. 78, a — b multiplied by c gives ac — bc, and multiplied
by d gives ad — bd. Subtracting the second result from the
first, we have
ac — bc — ad+bd
the product required.
80. We observe in the result of Art. 79,
1. The product of the positive term a by the positive term
c gives the positive term a c.
2. The product of the negative term —b by the positive
term c gives the negative term —be.
3. The product of the positive terra a by the negative term
— d gives the negative term — a d.
4. The product of the negative tern> —b by the negative
term — d gives the positive term b d.
26 ALGEBRA.
From these considerations we can state what is known as
the Rule of Signs in Multiplication, as follows:
+ multiplied by +, and — multiplied by — , produce + ;
+ multiplied by — , and — multiplied by + , produce — .
Or, as may he enunciated for the sake of hrevity with regard
to the product of any two terms,
Like signs produce + , and unlike signs produce — .
81. Let it he required to multiply 7 a by 2 b.
Since (Art. 77) the factors may he written in any order, we
have 7ax2b = 7x2xaXb = 14:ab. Hence,
The coefficient of the product is equal to the product of the
coefficients of the factors.
82. Let it he required to multiply a 3 hy a 2 .
By Art. 17, a s means sX«X« } and a 2 means aXa; hence,
a ! Xfl 2 = «X»X«X«X(i=« 5 . Hence,
The exponent of a letter in the product is equal to the sum
of its exponents in the factors. '
Or, in general, a m X a n = a m + n .
83. In Multiplication we may distinguish three cases.
CASE I. •
84. WJien both factors are monomials.
From Arts. 80, 81, and 82 is derived the following rule for
the product of any two monomials.
RULE.
Multiply the numerical coefficients together ; annex to the
residt the letters of both monomials, giving to curb letter an
exponent equal to the sum of its exponents in the factors. Make
the product + when the two factors have the same sign, and —
when they have different signs.
MULTIPLICATION. 27
EXAMPLES.
1. Multiply 2 «" by 3 a 2 .
2« 4 x3a 2 = 6a 6 , Ans.
2. Multiply a 3 b 2 c by  5 a 2 b d.
a 8 b 2 c X — 5 a 2 b d — — 5 a 5 b s c d, Ans.
3. Multiply — 7 x m by — 5 se n .
— 7 a; m x — 5 x n = 35 a: m+n , ^4?zs.
4. Multiply 3 a (a; — y) 2 by 4« 3 (x ?/).
3 a (a  y) 2 X 4 a 3 (xy) = 12 a 4 (a;  y) 3 , Ans.
Multiply the following :
5. 15 m 5 w 6 by 3 m n. 12. — 12 a 2 x by — 2 a 2 y.
6. 3 a 6 by 2 a e. 13. 3 a m 6 n by — 5 a n b r .
7. 17 a b c by — 8 a b c. 14. — 4 x m ?/" by — x n y n z b .
8.17 a 4 c 2 by  3 a 2 c 2 . 15. 2 a m 5" by 5 a 3 b.
9. 11 n 2 y by — 5 w 6 «. 16. — 7 m n x 2 by m n # r y 2 .
10. 4a 6 by3aiy 2 . 17. 2 m (a  b) 2 by m (a  b).
11. — 6 a b 2 c by a 3 b m. 18. 7 a (x — y) hy —3 a 2 b (x — y).
19. Find the continued product of 8 a x 2 , 2 a 3 y, and 4 a; 3 v/ 4 .
20. Find the continued product of 2 a c 2 , — 4 a c 3 , and
3«J 2 .
CASE II.
85. Wlien one of the factors is a polynomial.
From Art. 78 we have the following
RULE.
Multiply each term of the multiplicand by the multiplier,
remembering that like signs produce +, and unlike signs pro
duce — .
28 ALGEBRA.
EXAMPLES.
1. Multiply 3 x — y by 2 x y.
3 x — y
2 x y
6 x' 2 y — 2 x y 2 , Ans.
2. Multiply 3 a — 5 x by — 4 ra.
3 a — 5 x
— 4 m
,3
.2
— 12 a m + 20 ra #, ^4?is.
Multiply tbe following :
3. x 2 2z3by 4z. 7. x 4 10a; 3 + 5by2x
4. 8 a 2 6 c  fZ by 5 a f/ 2 . 8. a 2 + 13 a&  6 6 2 by 4 a b 2
5. 3 x 2 + 6 x — 7 by — 2 x s . 9. ra 2 + m n + ?r by m n.
6. 3 ra 2 — 5 ra ?i — ?i 2 by — 2 m. 10. 5 — 6 a — 8 a 3 by — 6 <x
11. 5a; 3 4x 2 3z2by6r\
12. « 3  3 a 2 b + 3 a b 2  b* by a 2 b 2 .
CASE III.
86. When both of the factors are polynomials.
In Art. 79 we sbowed that tbe product of a — b and c — d
rnigbt be obtained by multiplying a — b by c, and then by d,
and subtracting tbe second result from tbe first. It would
evidently be equally correct to multiply a — b by c, and then
by — d, and add the second result to the first. On this we
base the following rule for finding the product of two poly
nomials.
RULE.
Multiply each term of the multiplicand by each term of the
multiplier, remembering that like signs 'produce +, and unlike
signs produce —, and add the partial, products.
MULTIPLICATION. 29
EXAMPLES.
1. Multiply 3 a  2 b by 2 a 5b.
3a 2b
2a —5b
6 a 2 — Aab
 15 a b + 10 b 2
6 a 2 — 19 a b + 10 6' 2 , ^ws.
The reason for shifting the second partial product one place
to the right, is that in general it enables us to place like terms
in the same vertical column, where they are more conveniently
added.
2. Multiply x 2 + 1 — x 3 — x by x + 1.
1 — X + X 2 — X 3
1 + x
1 — X + X 2 — X 3
+ X — X 2 + X 3 — X*
1 —x 4 , Ans.
It is convenient, though not essential, to have both multi
plicand and multiplier arranged in the same order of powers
(Art. 41), and to write the product in the same order.
Multiply the following :
3. 3 x 2 — 2 x y — y 2 by 2 x — Ay.
4. x 2 + 2x + lhyx 2 2x + 3.
5. a + b — c by a — b + c.
6. 3a2bhy2a + 4.b.
7. a 2 + b 2 + ab by b — a.
8. 1 + x + x 3 + x 2 by a x — a.
9. 5 a 2  3 a b + 4 b 2 by 6 a  5 b.
10. 3 x 2 — 7 x + 4 by 2 x 2 + 9 x — 5.
30 ALGEBRA.
11. 6 x  2 x  5  a 3 by x 2 + 10  2 x.
12. 2a 3 +5a> 2 8a:7by45 : * ; 3a: 2 .
13. a 3 b  a 2 b 2  4 a 6 8 by 2 a 2 5  a b 2 .
14. x m + 2 ij — 3xy n ~ l by 4 *"• + 5 y 2 — 4 a 4 y n .
15. 6 a: 4  3 x 3  a: 2 +6 a; 2 by 2x 2 + x + 2.
16. m 4 — m 3 w + rn 2 n 2 — m w 3 + % 4 by m + n.
17. ft 3 3« 2 H3ai 2  6 3 by a 2 2ab + b\
87. It is sometimes sufficient to indicate the product of
polynomials, by enclosing each of the given factors in a paren
thesis, arid writing them one after the other, with or without
the sign X between the parentheses. When the indicated
multiplication is performed, the expression is said to be ex
panded or developed.
1. Indicate the product of 2 x 2 — 3 x y + 6 by 3 x 2 + 3 x y — o.
Kesult, (2x 2 3xy + 6) (3x 2 +3xy5).
EXAMPLES.
2. Expand (3 a + 4 b) (2 a + b).
3. Expand (a* — a 3 x + a 2 x 2 — a x 3 + x 4 ) (a + x).
4. Develop (a* — a: 4 ) X (« 4 — a,' 4 ).
5. Develop (a m — a") (2 a — a n ).
6. Expand (1 + x) (1 + a 4 ) (1 — a + .r 2  a 3 ).
7. Find the value of (« + 2 a) (« — 3 x) (a + 4 x).
8. Expand [« (a 2 — 3 a + 3)  1] x [a (a  2) + 1].
88. From the definition of Art. 76, X a means taken
a times. Since taken any number of times produces 0, it
follows that x a = 0. That is,
If zero be multiplied by any quantity} the product is equal
to zero.
DIVISION. 31
89. Since (+ a) X (+ b) —ab, and (— a) X (— &) =ab,
it follows that in the indicated product of two factors, all the
signs of both factors may be changed without altering the value
of the expression. Thus,
(x — y) {a — b) is equal to {y — x) (J> — a).
Similarly we may show that in the indicated product of any
number of factors, any even number of factors may In/re their
signs changed without altering the value of the expression.
Thus, (x — y) (c — d) (e —f) (g — h) is equal to
{yx) (cd) (fe) (g h), or to
(y — x) (d — c) {fe) (h — g), etc.; but is not equal to
(yx){d~c)(fe)(gh).
VI. — DIVISION.
90. Division, in Algebra, is the process of finding one of
two factors, when their product and the other factor are given.
Hence, Division is the converse of Multiplication.
The Dividend is the product of the two factors.
The Divisor is the given factor.
The Quotient is the required factor.
91. Since the quotient multiplied by the divisor produces
the dividend, it follows, from Art. 80, that if the divisor and
quotient have the same sign, the dividend is + ; and if they
have different signs, the dividend is — . Hence,
+ divided by +, and — divided by — , produce + ;
+ divided by — , and — divided by +, produce — .
Hence, in division as in multiplication,
Like signs produce +, and unlike signs produce — .
32 ALGEBRA.
92. Let it be required to find the quotient of 14 a b divided
by 7 a.
Since the quotient is such a quantity as when multiplied by
the divisor produces the dividend, the quotient required must
be such a quantity as when multiplied by 7 a will produce
14 a b. That quantity is evidently 2 b. Hence,
The coefficient of the quotient is equal to the coefficient of
the dividend divided by the coefficient of the divisor.
93. Let it be required to find the quotient of a 5 divided
by a 3 .
The quotient required must be such a quantity as when
multiplied by a 3 will produce a 5 . That quantity is evidently
a 2 . Hence,
The exponent of a letter in the quotient is equal to its expo
nent in the dividend diminished by its exponent in the divisor.
Or, in general, a m f a n = a m ~ n .
94. If we apply the rule of Art. 93 to finding the quotient
of a™ divided by a m , we have a m f a m = a m ~ m = a .
Now, according to the previously given definition of an ex
ponent (Art. 17), a° has no meaning, and we are therefore at
liberty to give to it any definition we please. As a m — a m = 1,
we should naturally define a° as being equal to 1 ; and as a
may represent any quantity whatever,
Any quantity whose exponent is is' equal to 1.
By this notation, the trace of a letter which has disappeared
in the operation of division may be preserved. Thus, the
quotient of a 2 b 3 divided by a 2 b 2 , if important to indicate that
a originally entered into the term, may be written a b.
95. In Division we may distinguish three cases.
CASE I.
96. When both dividend and divisor are monomials.
From the preceding articles is derived the following
DIVISION. 33
RULE.
Divide the coefficient of the dividend by that of the divisor ;
and to the result annex the letters of the dividend, each with an
exponent equal to its exponent in the dividend diminished by
its exponent in the divisor; omitting all letters whose expo
nents become zero. Make the quotient + when the dividend and
divisor have the same sign, and — when they have different
signs.
EXAMPLES.
1. Divide 9 a 2 b c x y by 3 a b c.
9a 2 bcxyi3abc = 3axy, Ans.
2. Divide 24 a 4 m 3 n 2 by — 8 a m 3 n.
24 a 4 m 3 n 2 . — 8 a m 3 n — — 3 a 3 n, Ans.
3. Divide — 35 x m by — 7 x n .
— 35 x m ^ — 7x n = 5 x m ~ n , Ans.
Divide tbe following :
4. 12 a 5 by 4 a. 8.  65 a 3 b 3 c 3 by  5 a b 2 c 3 .
5. 6 a 2 c by 6 a c. 9. 72 m° n by — 12 m 2 .
6. 14 m 3 n 4 by  7 m n 3 . 10.  144 c 5 <P e 6 by 36 c 2 d 3 e.
7. 18x 2 y 5 zhy9x 2 z. 11.  91 x 4 y 3 z 2 by  13 x 3 y 1 .
CASE II.
97. When the dividend is a polynomial and the divisor is
a monomial.
Tbe operation being just tbe reverse of that of Art. 85, we
have the following
RULE.
Divide each term of the dividend by the divisor, remembering
that like signs produce +, and tinlike signs produce — .
34 ALGEBRA.
EXAMPLES.
1. Divide 9 a 3 ft + 6 a 4 e — 12 a ft by 3 a.
3«)9a 3 H6 a 4 c — 12 a 5
3a 2 i + 2« 3 c4J, ^4»s.
Divide the following :
2. 8 a 3 6 c + 16 a 5 6 c — 4 a 2 c 2 by 4 a 2 c.
3. 9 a 5 ft c  3 a 2 ft + 18 a 3 ft c by 3 a ft.
4. 20 a 4 ft c + 15 a 6 d 3  10 a 2 ft by  5 a ft.
5. 3 a 3 (a  ft) + 9 a (a + ft) by 3 a.
6. 15 (x + y) 2 — 5 a (x + y) + 10 ft (x + y) by — 5 (x + y).
7. 4 x 7  8 a 6  14 t> + 2 a 4  6 x 3 by 2 x\
8. 9 a 4 + 27 x 3  21 a 2 by 3 a; 2 .
9. _ a 6 £6 C 4 _ a 4 &5 c 3 + 3 a 8 £4 ^2 fcy _ ft 8 p ^
10. — 12 aP ft?  30 a 12 ft 3 + 108 a" ft n by  6 a m ft'".
CASE III.
98. When the divisor is a polynomial.
1. Let it be required to divide 12 + 10 x 3 — 11 x — 21 x 2 by
2z 2 43x.
We are then to find a quantity which when multiplied by
2 x 2  4  3 x will produce 12 + 10 x 3  11 x  21 x 2 .
Now, in the product of two polynomials, the term containing
the highest power of any letter in the multiplicand, multiplied
by the term containing the highest power of the same letter
in the multiplier, produces the term containing the highest
power of that letter in the product. Hence, if the term con
taining the highest power of x in the dividend, 10 x 3 , be di
vided by the term containing the highest power of x in the
divisor, 2 x 2 , the result, 5 x, will he the term containing the
highest power of x in the quotient.
DIVISION. 35
Multiplying the divisor by 5 x, the term of the quotient
already found, and subtracting the result, 10 x 3 — 20 x — 15 x 2 ,
from the dividend,' the remainder, 12 + 9 x — 6 x' 2 , may be re
garded as the product of the divisor by the rest of the quotient.
Therefore, to find the rest of the quotient, we proceed as be
fore, regarding 12 + 9 x — 6 x 2 as a new dividend, and divid
ing the term containing the highest power of x, — 6 x' 2 , by
the term containing the highest power of x in the divisor, 2 x 2 ,
giving as a result — 3, which is the term containing the high
est power of x in the rest of the quotient.
Multiplying the divisor by — 3, the term of the quotient
just found, and subtracting the result, — 6 x 2 + 12 + 9 x, from
the second dividend, there is no remainder. Hence, 5 x — 3
is the quotient required.
99. It will be observed that in getting the terms of the
quotient, we search for the terms containing the highest power
of some letter in the dividend and divisor. These may be
obtained most conveniently by arranging both dividend and
divisor in order of powers commencing with the highest
(Art. 41) ; this, too, facilitates the subsequent subtraction.
We also should arrange each remainder or new dividend in
the same order.
It is customary to arrange the work as follows :
10 x 3  21 x 2  11 x + 12
10 x s  15 x 2  20 x
2 x 2 — 3 x — 4, Divisor.
5 x — 3, Quotient.
— 6x 2 + 9 a: + 12
 6x 2 + 9 a; + 12
100. We might have obtained the quotient by dividing the
term containing the lowest power of x in the dividend, 12, by
the term containing the lowest power of x in the divisor, — 4,
which would have given as a result — 3, the term containing
the lowest power of x in the quotient. In solving the problem
in this way, we should first arrange both dividend and divisor
in order of powers commencing with the lowest, and should
o
G ALGEBRA.
afterwards bring clown each remainder in the same order; re
membering that a term which does not contain x at all con
tains a lower power of x than any term which contains x.
101. From the preceding principles we derive the follow
ing
RULE.
Arrange both dividend and divisor in the same order of pow
ers of some common letter.
Divide the first term of the dividend by the first term of the
divisor, and write the result as the first term of the quotient.
Multiply the whole divisor by this term, and subtract the
product from the dividend, arranging the result in the same
order of powers as the divisor and dividend.
Regard the remainder as a new dividend, and divide its first
term by the first term of the divisor, giving the next term of the
quotient.
Multiply the whole divisor by this term, and subtract the
product from the last remainder.
Continue in the same manner until the remainder becomes
zero, or until the first term of the remainder will not contain
the first term of the divisor.
When a remainder is found whose first term will not con
tain the first term of the divisor, the remainder may be written
with the divisor under it in the form of a fraction, and added
to the quotient.
2. Divide a 8  3 a 2 b +. 12 b z + 5 a b 2 by b + a.
Arranging the dividend and divisor in order of powers,
a + b) a 3  3 a 2 b + 5 a b 2 + 12 b z {a 2  4 a b + 9 b 2
a 3 + a 2 b
 4 a 2 b ~
— 4 a 2 b — 4 a b 2
9ab 2
9 a b 2 +9b s
3 b 3 , Remainder.
3 b 9
Ans, a 2 4ab + 9b 2 +
a + b'
DIVISION. 37
EXAMPLES.
3. Divide 2 a 2 x 2 — 5 a x + 2 by 2 a x — 1.
4. Divide 3 6 3 + 3 a b 2  4 a 2 b  4 « 3 by a + b.
5. Divide 8 a?  4 a 2 i  6 a b 2 + 3 & 3 by 2 a  J.
6. Divide 21 a 5  21 b 5 by la — lb.
7. Divide a 3 + 2 a; 3 by a + #.
8. Divide x* + y 4 by cc + y.
9. Divide 23 x" 48 + 6 cc 4  2 a;  31 x* by 6 + 3a; 2 — 5 x.
10. Divide 15 x 4  32 x 3 + 50 x 2  32 a; + 15 by 3 x 2  4 x + 5.
11. Divide 2 a; 4  11 aj  4 ar  12  3 x 3 by 4 + 2 ar + jb.
12. Divide a; 5 — v/ 5 by x — y.
13. Divide 35  17 x + 16 x 2  25 a; 3 + 6 x 4 by 2 x — 1.
14. Divide 3 x 2 + 4 x + 6 a;  11 x 3  4 by 3 x 2  4.
15. Divide a 2 — &' 2 + 2 & c — c 2 by a + b — c.
16. Divide a; 4 — 9 a; 2 — 6 x y — y 2 by x 2 + 3 x + y.
17. Divide x n + 1 + x n y + x y n + y n + 1 by x n + y n .
18. Divide cr n — b 2m + 2 b m c r — c 2r by a n + b m — c r .
19. Divide 1 + a hy 1 — a.
In examples of this kind the division does not terminate,
there being a remainder however far the operation may be
carried.
20. Divide a by 1 + x.
21. Divide a 8 + a 6 b 2 + a 4 b 4 + a 2 b* + b a
by a 4 +a 3 b + a 2 b 2 + ab 3 + b 4 .
22. Divide 3 a 3 + 2  4 a 5 + 7 a + 2 a 6  5 a 4 + 10 a 2
by a 3 — 1 — a 2 — 2 a.
23. Divide 15 x 2  x 4  20  2 a; 5 + 6 x + 2 x 3
by 5  3 a; 2  4 x + 2 x 3 .
38 ALGEBRA.
24. Divide 2 x 5 + 4 x 2 — 14 + 7 x — 7 x 3 + x« — x i
by 2 x 2  7 + z 3 .
25. Divide 12 « 5  14 a 4 b + 10 a 8 i 2  a 2 6 3  8 a i 4 + 4 i 5
by 6 a 8  4 a 2 6  3 a b 2 + 2 6 3 .
102. In Art. 88 we showed that X a = 0. Since the
product of the divisor and quotient equals the dividend, we
may regard as the quotient, a as the divisor, and as the
dividend. Therefore,
° = 0.
That is, a
If zero be divided by any quantity the quotient is equal to
zero.
VII. — FORMULAE.
103. A Formula is an algebraic expression of a general rule.
The following formulae will be found very useful in abridg
ing algebraic operations.
104. By Art. 17, (a + bf =5 (a + b) (a + b) ; whence, by
actual multiplication, we have
That is, ( a + V* = cc 2 + 2ab + b 2 . (1)
The square of the sum of two quantities is equal to the
square of the first, plus twice the product of the first by the
second, plus the square of the second.
105. We may also show, by multiplication, that
(a  b) 2 = a 2 2ab + b\ (2)
That is,
The square of the difference of two quantities is equal to
the square of the first, minus twice the product of the first by
the second, plus the square of the second.
106. Again, by multiplication, we have
(a + b) (ab) = a 2  b 2 . (3)
FORMULAE. 3y
That is,
The product of the sum and difference of two quantities is
equal to the difference of their squares.
EXAMPLES.
1 107. 1. Square 3 a + 2 b.
The square of the first term is 9 a 2 , twice the product of the
terms is 12 a b, and the square of the last term is 4 b 2 . Hence,
by formula (1),
(3 a + 2 b) 2 = 9 a 2 + 12 a b + 4 b 2 , Am.
Square the following :
2. 2m + 3 w. 4. 3 x + 11. 6. 2ab + &ac.
3. x 2 + 4. 5. 4a + 5 b. 7. 7 x 3 +3x.
8. Square 4 x — 5.
The square of the first term is 16 x 2 , twice the product of
the terms is 40 a, and the square of the last term is 25.
Hence, by formula (2),
(4 x  5) 2 = 16 x 2  40 x + 25, Am.
Square the following :
9. 3a 2 b s . 11. l2£ 13. 3a\
10. 4 a b  x. 12. x*  y\ 14. 2 a; 3  9 x 2 .
15. Multiply 6 a + b by 6 « — 6.
The square, of the first term is 36 a 2 , and of the last term b 2 .
Hence, by formula (3),
(6a + b) (6 a  b) = 36 a 2  b 2 , Am.
Expand the following :
16. (x + 3) (x  3). 19. (a m + a") (a m  a n ).
17. (2 x + 1) (2 x  1). 20. O 3 + 5 a) (a; 3  5 a).
18. (5« + 7J) (5a76). 21. (4ar + 3) (4a; 2 3).
22. Multiply a + b + chja + b — c.
(a + b + c) (a + b  c) = [(» + b) + c] [O + 6)  c]
= (Art. 106), (a + b) 2  c 2 = a 2 + 2 « b + b 2  c 2 , Am.
40 ALGEBRA.
Expand the following :
23. [1 + («&)] [1 (« &)]• 25  (a5 + c)(a — &c).
24. (a + & + c) (a — & — c). 26. (c + a  &) (c — a + b).
27. [(a + b) + (cd)2 [(a + ft)  (««*)].
28. (a — b + c — d)(a — b — c + d).
29. (a + b + c + d) (a + b — c — d).
VIII. — FACTORING.
108. The Factors of a quantity are such quantities as will
divide it without a remainder.
109. Factoring is the process of resolving a quantity into
its factors.
110. A Prime Quantity is one that cannot be divided,
without a remainder, by any integral quantity, except itself
or unity.
Thus, a, b, and a + c are prime quantities.
Quantities are said to he prime to each other when they have
no common integral divisor except unity.
111. One quantity is said to be divisible by another when
the latter will divide the former without a remainder.
Thus, a b and a b + a 2 b' 1 are both divisible by a, b, and a b.
112. If a quantity can be resolved into two equal factors,
it is said to be a, perfect square ; and one of the equal factors
is called the square root of the quantity.
If a quantity can be resolved into three equal factors, it is
said to be a perfect cube ; and one of the equal factors is called
the cube root of the quantity.
Thus, since 1 a equals 2 a X 2 a, 4 a 2 is a perfect square
and 2 a is its square root ; and since 27 X s equals 3,xx3a , x3/,
27 X s is a perfect cube, and 3x is its <nbe root.
FACTORING. 41
Note. 4 a 2 also equals  2 a x  2 a, so that the square root of 4 <> 2
may be either 2 a or  2 «. In the examples in this chapter we shall only
consider the positive square root.
To find the square root of an algebraic quantity, extract the square root
of the numerical coefficient, and divide the exponent of each letter by 2.
Thus, the square root of 9 a 6 b 2 is 3 a? b.
To find the cube root, extract the cube root of the numerical coefficient,
and divide the exponent of each letter by 3. Thus, the cube root of 8 a 3 b®
is 2 a b 2 .
113. The factoring of monomials may be performed by
inspection ; thus,
12 a 3 b 2 c = 2.2.8. a a abbe.
But in the decomposition of polynomials we are governed by
rules which may be derived from the laws of their formation.
A polynomial is not always factorable ; but in numerous cases
we can always factor ; and these cases, together with the rules
for their solution, will be found in the succeeding articles.
CASE I.
114. Wlien the terms of a polynomial have a common mo
nomial factor, it may be written as one of the factors of the
polynomial, ivith the quotient obtained by dividing the given
polynomial by this factor, as the other.
1. Factor the expression 3 x 3 y 2 — 12 x y 4 — 9 x 2 y 3 .
We observe that each term contains the factor 3 x y 2 .
Dividing the given polynomial by 3 x y 2 , we obtain as a
quotient x 2 — 4 y 2 — 3 x y. Hence,
3 x 3 y 2  12 x y 4  9 x 2 y z = 3 .« y 2 (x 2 ±y 2 3x y), Arts.
EXAMPLES.
Factor the following expressions :
2. a s + a. $ 5. 60m 4 n 2 — 12 n\
3. 16 x 4  12 x. 6. 27 c 4 d 2 + 9 c 3 d.
4. a & 2 a 4 + 3 a 3 a 2 . 7. 36 X s y — 60 x 2 y 4 — 84 x 4 y 2 .
8. a 5 b3a 6 b 4 2a 3 b 4 c + 6a' 1 b 5 x.
9. 84 x 2 y 3  140 x 3 y 4 + 56 x 4 if.
42 ALGEBRA.
10. 72 o 4 b 3 c 3 + 126 a 3 c 2 d + 162 a 2 c.
1 1. 128 c 4 d 5 + 320 c 2 d 7  448 c 8 <Z 4 .
CASE II.
115. TFAerc a polynomial consists of four terms, the first
two and last two of which have a common binomial factor, it
may he written as one of the factors of the polynomial, wit li
the quotient obtained by dividing the given polynomial by this
factor, as the other.
1. Factor the expression a m — b m + a • n — b n.
Factoring the first two and last two terms by the method
of Case I, we obtain m (a — b) + n (a — b). We observe that
the first two and last two terms have the common binomial
factor a — b. Dividing the expression by this, we obtain as a
quotient m + n.
Hence, am — bm + an — bn=(a — b) (m + n), Ans.
2. Factor the expression a m — bm — an + b n.
am — b 7)i — a n + b n =■ a m — b m — (a n — b 11) = ni (a — b)
— n(a — b) — (a — b) (m — n), Ans.
Note. If the third term of the four is negative, as in Ex. 2, it is
convenient to enclose the last two terms in a parenthesis with a  sign
prefixed, before factoring.
EXAMPLES.
Factor the following expressions :
3. a b + b x + a y + x y. 7. mx 2 — my 2 — ?ix 2 + n y 1 .
4. a c — cm + a d — dm. 8. x 3 + x 2 + x + 1.
5. x 2 + 2x — xy — 2y. 9. 6 x 3 + 4 x 2 — 9 x  6.
6. a 3 — a 2 b + a b 2  b 3 . 10. 8 c x  12 c y + 2 d x  3 d y.
11. 6 n  21 m 2 n8m + 28 m 3 .
12. a 2 bc — ac 2 d+ab 2 d — bc d 2 .
13. m 2 n 2 x 2 — n s x y — m 3 x y + m n y' 2 .
14. 12 a b m n — 21 a b x y + 20 <■ d m n — 35 c d x //.
FACTORING. 4:3
CASE III.
116. When the first and last terms of a trinomial are
perfect squares and positive, and the second term is twice the
product of their square roots.
Comparing with Formulae 1 and 2, Arts. 104 and 105, we
observe that such expressions are produced by the product of
two equal binomial factors. Reversing the rules of Arts. 104
and 105, we have the following rule for obtaining one of the
equal factors :
Extract the square roots of the first and last terms, and
connect the results by the sign of the second term.
1. Factor a 2 + 2 a b + b 2 .
The square root of the first term is a ; of the last term, b ;
the sign of the second term is + . Hence, one of the equal
factors is a + b.
Therefore, a 2 + 2 a b + b 2 = (a + b) (a + b) or (a + b) 2 , Ans.
2. Factor 9 or — 12 a, b + A hi
The square root of the first term is 3 a ; of the last term, 2 b ;
the sign of the second term is — . Hence, one of the equal
factors' is 3 a — 2 b. Therefore,
9 a 2  12 a b + 4 b 2 = (3 a 2 b) (3 a  2 b) or (3 a  2 bf, Ans.
Note. According to Art. 58, the given expression may be written
4 b 2 — 12 a b + 9 a 2 . Applying the rule to this expression, we have
4 b 2  12 a b + 9 a 2 = (2 b 3 a) (2 b  3 a) or (2 b  3 a) 2 .
We should obtain this second form of the result in another way by apply
ing the principles of Art. 89 to the first factors obtained.
EXAMPLES.
Factor the following expressions :
3. x 2  14 x + 49. 6. a 2  28 a + 196..
4. m 2 + 36 m + 324. 7. n 6  26 n 3 + 169.
5. y 2 + 20 y + 100. 8. x 2 y 2 + 32 x y + 256.
44 ALGEBRA.
9. 25 x 2 + 70 x y z + 49 y 2 z 2 . 11. 16 m 2 8am+« 2 .
10. 36 m?  36 w « + 9 n 2 . 12. 4 a 2 + 44 a b + 121 b 2 .
13. a 2 6 4 + 12 a b 2 c + 36 c 2 .
14. 9 « 4 + 60 a 2 bc 2 d + 100 b 2 c i d 2 .
15. 4 x A  60 m re x 2 + 225 m 2 t»s
16. 64 x 6  160 x 5 + 100 cc 4 .
CASE IV.
117. When an expression is the difference between two
perfect squares.
Comparing with Formula 3, Art. 106, we observe that such
expressions are the product of the sum and difference of two
quantities. Reversing the rule of Art. 106, we have the fol
lowing rule for obtaining the factors :
Extract the square roots of the first and last terms ; add
the results for one factor, and subtract the second result from
the first for the other. •
1. Factor 36 x 2 — 49 y 2 .
The square root of the first term is 6 a;; of the last, 7 y.
The sum of these is 6 x + 7 y, and the second subtracted from
the first is 6 x — 7 y. Hence,
36 x 2 — 49 y 2 = (6 x + 7 y) (6 x — 7 y), Ans.
2. Factor {a  b) 2  (cdf.
The square root of the first term is a — b ; of the last, c — d.
The sum of these is a — b + c — d, and the second subtracted
from the first is a — b — c + d. Hence,
(a — b) 2 — (c — d) 2 = (a — b + c — d) (a — b — c + cl), Ans.
EXAMPLES.
Factor the following expressions :
3. x 2 l. 5. a 4 /A 7. 4 a* — 225 m a w 9 .
4. 4x 2 9t/ 2 . 6. 9 a 2 4. 8. 1  196 x 2 if z\
FACTORING. 45
9. (a + b) 2  (c + d) 2 . 11. m 2  (x  y) 2 .
10. (ac)*—b*. 12. (aj — m) a — (y — »)*.
Many polynomials, consisting of four or six terms, may be
expressed as the difference between two perfect squares, and
hence may be factored by the rule of Case IV.
13. Factor 2 m n + m 2 — 1 + n 2 .
Arrange the expression as follows, m 2 + 2mn + r? — 1.
Applying the method of .Case III to the first three terms, we
may write the expression (m + n)' 2 — 1. The square root of
the first term is m + n ; of the last, 1. The sum of these is
m + n + 1, and the second subtracted from the first is
m + n — 1. Hence,
2 m n + m 2 — 1 + n 2 = (m + n + 1) (m + n — 1), Ans.
14. Factor 2 x y + 1 — x 2 — y 2 .
2 x y + 1 — x 2 — y 2 = l — x 2 \2x y — y 2
= 1 {x 2 2 x y + y 2 ) = l  (xy) 2 , by Case III,
= [1 + (x— y)][l — (as— y)] = (l + #y) (1— x + y),Ans.
15. Factor 2xy+b 2 x 2 2ab y 2 + a 2 .
2xy+b 2 — x 2 — 2ab — y 2 + a 2
= a 2 — 2ab + b' 2 — x 2 + 2 x y — y 2
= a 2 2 a b + b 2  (x s  2 x y + y 2 )
= (a  b) 2 — (x— y) 2 , by Case III,
= [(«&)+ (asy)][(afl)(ajy)]
= (a — b+x — y) (a—b — x + y), Ans.
Factor the following expressions :
16. x 2 + 2xy + y 2 4:. 19. 9x 4 4?/ 2 + 4f 2 y.
17. a a _ j2 + 2 h c _ c 2. 20. 4 a 2 + 6 2  9 d? 4ab.
18. 9 c 2  1 + d? + 6 c d. 21. 4 b  1  4 b 2 + 4 m 4 .
22. a 2 — 2 a m + m 2 — b 2 — 2 b n — n 2 .
23. 2 a m — 6 2 + m 2 + 2bn+ a 2  n 2 .
24. x 2  y 2 + c 2  d 2  2 c x + 2 tf y.
46 ALGEBRA.
CASE V.
118. Wlien an expression is a trinomial, of the form
x + a x + b ; where the coefficient of x 2 is unity, and a and
b represent any whole numbers, either positive or negative.
To derive a rule for this case we will consider four examples
in Multiplication :
I. ii.
x + 5 x — o
x + 3 x — 3
x 1 + 5 x x 2 —ox
+ 3.x + 15 3cc + 15
a; 2 + Sx + 15 sc a 8aT+15
III. IV.
x + 5 x — 5
x — 3 x + 3
a? 2 + 5 x # 2 — 5 .<■
3x15 +3xll
« 2 +2^15 x 2 2ic15
We observe in these results,
1. The coefficient of x is the algebraic sum of the numbers
in the factors.
2. The last term is the product of the numbers.
Hence, in reversing the process, we have the following rule
for obtaining the numbers :
RULE.
Find two numbers ivhose algebraic sum is the coefficient of
x, and whose product is the last term.
Note. We may shorten the work by considering the following points :
1. When the last term of the product is f, as in Examples I ami II,
the sum of the numbers is the coefficient of .<• ; both numbers being +
when the second term is +, and  when the second term is .
2. When the last term is , as in Examples III and I V, the difference
FACTORING. 47
of the numbers (disregarding signs) is the coefficient of x ; the larger
number being + and the smaller  when the second term is +, and the
larger number  and the smaller + when the second term is  .
We may embody these observations in the form of a rule which may be
found more convenient than the preceding rule in the solution of examples.
I. If the last term is +, find tivo members whose sum is the coefficient of
x, and whose product is the last term; and give to both numbers the sign of
the second term.
II. If the last term is  , find tivo numbers whose difference is the coeffi
cient of x, and whose product is the last term; and give to the larger num
ber the sign of the second term, and to the smaller number tlie opposite sign.
1. Factor x 2 + 14 x + 45.
Here we are to find two numbers whose  " 
(.product — 45 J
The numbers are 9 and 5 ; and, the second term being + , both
numbers are +. Hence,
x 2 + 14 x + 45 = (x + 9) (x + 5), Ans.
2. Factor x 2 — 6 x + 5.
Here we are to find two numbers whose ■! \
(product = 5,'
The numbers are 5 and 1 ; and, as the second term is — , both
numbers are — . Hence,
x 2 — 6 x + 5 = (x — 5) (x — 1), Ans.
3. Factor x 2 + 5 x — 14.
Here we are to find two numbers whose ■< "
I product = 14 )
The numbers are 7 and 2; and as the second term is +, the
larger number is + , and the smaller — . Hence,
x 2 + 5 x — 14 = (x + 7) (x — 2), Ans.
4. Factor x 2 — 7 x — 30.
Here we are to find two numbers whose ] ~ \
48 ALGEBKA.
The numbers are 10 and 3; and as the second term is — , the
larger number is — , and the smaller + . Hence,
x 2  7 x  30 = (x  10) (x + 3), Arts.
Note. In case the numbers cannot be readily determined by inspection,
the following method will always give them :
Eequired two numbers whose difference is 8 and product 48. Taking in
order, beginning with the lowest, all possible pairs of integral factors of 48,
we have
1x48,
2x24,
3x16,
4x12.
And, as 4 and 1 2 differ by 8, they are the numbers required.
Evidently this method will give the required numbers eventually, how
ever large they may be, provided they exist.
EXAMPLES.
Factor the following expressions :
5. a 2 +5 a + 6. 12. m 2 +9m + 8.
6. a 2 3a + 2. 13. m' 2 + 2m80.
7. 2/2 + 2?/ 8. 14. c 2  18 c + 32.
8. m 2 5m24. 15. x 2 + x42.
9. * 2 ll;c + l8. 16. x 2 + 23x + 102.
10. n * n c)0. 17. if9y90.
11. * 2 +13a; + 36. 18. a 2 +13a48.
19. cc 2 9z70.
20. Factor 15 — 2x — x 2 .
15 _ 2 x  x 2 =  (x 2 + 2 x  15)
By the rule of Case V, x 2 + 2 x  15 = (.r + 5) (x  3).
Hence,
15  2 x  x 2 =  (x + 5) (x  3) = (x + 5) (3  x), Ans.
Note. If the x* term is , enclose the whole expression in a paren
thesis with a  sign prefixed. Factor the quantity within the parenthesis,
and change the signs of all the terms of one factor.
FACTORING. 49
Factor the following expressions :
21. 20xx 2 . 22. 6 + xx 2 . 23. 845za; 2 .
The method of Case V may he extended to the factoring of
more complicated trinomials.
24. Factor m 2 ri 2 — 3 m nx + 2 x 2 .
r. i it (sum =3)
Here we are to find two numbers whose < t — 9 j
The numbers are 2 and 1; and as the second term is—,
hoth numbers are — . Hence,
m 2 ri 2 — 3 m nx + 2x 2 = (m n — 2 x) (m n — x), Ans.
Factor the following expressions :
25. a 4  29 x 2 + 120. 30. ?« 4 + 5 m 2 n 2  66 n\
26. c 6 + 12c 3 +ll. 31. (ab) 2 S(ab)4.
27. x 2 y G + 2xf120. 32. (x + y)* 7 (x + y) + 10.
28. (r '^_7«i 2 144. 33. x 2  2 x y 2 z  48 y* z\
29. x 2 + 25 w x + 100 ?i 2 . 34. (m + nf + (m + n)  2.
CASE VI.
119. When an expression is the sum or difference of two
perfect cubes.
By actual division, we may show that
a 3 + b 3 a 3 — b 3
— — = a 2 — ab + b' 2 , and — = a 2 + a b + b 2 .
a + b a — b
Whence,
(a 3 + b 3 ) = (a + b) (a 2 ab + b 2 ), and
(a 3  b 3 ) = (a  b) (a 2 + ab + b 2 ).
These results may he enunciated as follows :
To factor the sum of two perfect cubes, write for the first
factor the sum of the cube 7'oots of the quantities; and for the
50 ALGEBEA.
second, the square of the first term of the first factor, minus
the product of the two terms, phis the square of the last term.
To factor the difference of two perfect cubes, write for the
first factor the difference of the cube roots of the quantities ;
and for the second, the square of the first term of the first
factor, plus the product of the two terms, plus the square of
the last term.
1. Factor 8 a 3 + 1.
The cube root of the first term is 2 a ; of the last term, 1.
Hence, 8 a 3 + 1 = (2 'a + 1) (4 a 2  2 a + 1), Ans.
2. Factor 27 x 6  64 y 3 .
The cube root of the first term is 3 x 2 ; of the last term, 4 y.
Hence,
27 x G  64 y* = (3 x 2  4 y) (9 x i + 12 x 2 y + 16 y 2 ), Ans.
EXAMPLES.
Factor the following expressions :
6. Sc 6 d 9 . 9. 343 + 8 a 3 .
7. 125 a s  216 m 3 . 10. 27 a; 3 125.
8. 729 c 3 dP + 512. 11. 1000 27 a 3 b\
CASE VII.
120. When an expression is the sum or difference of two
like powers ofttvo quantities. ■
The following principles are useful to remember :
1. a n — b n is always divisible by a — b, if n is an. integer.
2. a n — b n is always divisible by a + b, if n is an even integer.
3. a n + b n is always divisible by a + b, if n is an odd integer.
We may prove the first principle as follows :
Commencing the division of a n — b n by a — b, wv have
3.
iC 3 — y 3 .
4.
a 3 + 8.
5.
m 3 + 64 ?z, 6 .
a" — b n
FACTORING. 51
a — b
a n ' + . . . Quotient.

a n ] b — b n Remainder.
a n — b n , a n  } b — b n , b (a n ~ l — b n ~ v )
or,  —  =a n ~ i H ; = a n ~ l \ ',
a — b a — b a — b
It is evident from this result that, if a"" 1 — & n_1 is exactly
divisible by a — b, the dividend a n — b n will be exactly divisi
ble by a, — b. That is, if the difference of two like powers of
two quantities is divisible by the difference of those quantities,
then the difference of the next higher powers of the same
quantities is also divisible by the difference of the quantities.
But a s — b 3 is divisible by a — b, hence a 4 — b 4 is ; and since
c^ — b* is divisible by a — b, a 5 — b 5 is; and so on to any
power. This proves the first principle.
Similarly the second and third principles may be proved.
By continuing the division, we should find,
= a" 1 + a n ~ 2 b + a n ~ 3 b~ + + ab n ~ 2 + b n  1 (1)
a n ~' 2 b + a n  s b 2 — + ab n ~ 2 — fi"" 1 ' (2)
a
b
10
a n
a
— b n
+ b
=
a n 
l
a n
L.
+ b n
—
a n ~
l
a + b
a
2 b + a n  3 b 2  —ab n  2 +b n ~ 1 (3)
It is useful to remember that when a — b is the divisor, all
the terms of the quotient are + ; where a + b is the divisor,
the terms of the quotient are alternately + and — , the last
term being + if n is odd, and — if n is even.,
1. Factor a 1 — IP.
Putting n — 7 in (1), we have
a 1 b
a* + a 5 b + a 4 b 2 + a 3 b 3 + a 2 b* + ab s + b 6 .
a — b
Hence,
a 7  V = (a  b) (a 6 + a 5 b + a 4 b 2 + a 3 b 3 + a 2 b 4 + a b 5 + b 6 ),
Ans.
52 ALGEBRA.
2. Factor m 5 + x 5 .
Putting a = m, b = x, n== 5, in (3), we have
1YI f X . „ 2 2 3,4
= m — m° x + m x — mx A + x\
m + x
Hence,
m 5 + x s = (m + x) (ra 4 — m 3 a; + ra 2 a: 2 — m x 3 + x 4 ), Ans.
3. Factor x 6 — y 6 .
Putting a = x, b = y, n = 6, in (1), we have
x 6 ifi
— = x s + x* y + x s y 2 + x 2 y 3 + x y* + y 5 .
Hence,
x 6 — y 6 = (x — y) (x 5 + x 4 y + x s y 2 + x 2 y s + x y* + f), Ans.
Or, putting a =x, b = y, ?i = 6, in (2), we have
x 6 ?/ 6
J_ — ^5 ~,i », I ~,3 „fi ™2 „.3 i_ ™ ,.4 „.5
Hence,
= x° — a;* ?/ + x" y — x" y 6 + x if — y"
x 6 — y 6 =(x + y) (x h — x 4 y + x s y 2 — x 2 y 3 + xy i — y s ), Ans.
EXAMPLES.
Factor the following expressions :
4. x 5 + y 5 . 6. to 6 c 6 . 8. m s n*. 10. a 4 16.
5. (*d\ 7. a' + V. 9. c 7 l. 11. a 7 +128.
121. By the application of one or more of the given rules
for factoring, a quantity may sometimes be separated into
more than two factors.
1. Factor 2 a x s y 2  8 a x y\
By Case I, 2 a x 3 if  8 a x if = 2 a x y 2 (x 2  4 if).
Factoring the quantity in the parenthesis by Case IV,
2 a x 3 if — 8 a x if = 2 a x y 2 (x + 2 y) (x — 2 y), Ans.
Note. If the method of Case I is to be used in connection with other
cases, it should be applied first.
GREATEST COMMON DIVISOR. 53
2. Resolve a G — b 6 into four factors.
By Case IV, a 6  b G = (a 3 + b 3 ) (a 3  b 3 ).
By Case VI, a 3 + b 3 = (a + b) (a 2 ab + b 2 ),
and a 3  b 3 = (a  b) (a 2 + ab + b 2 ).
Hence,
a 6 b G =(a + b) (a b) (a 2 ab + b 2 ) (a 2 +ab + b 2 ), Ans.
EXAMPLES.
Factor the following expressions :
3. 3 a 3 b + 12 a 2 b + 12 a b. 7. 3 « 4  21 a 3 + 30 a:
4. 45 x 3 if — 120 x 2 f + 80 x if. 8. 2 c 3 m + 8 c 2 ra42 c m.
5. 18 x 3 y — 2 x f. 9. m 2 a;y4wa;y— 12 ay.
6. x 3 + Sx 2 +7x. 10. 32 « 4 b + 4 a 6 4 .
11. Resolve n 9 — 1 into three factors.
12. Resolve x i — if into three factors.
13. Resolve x 8 — m 8 into four factors.
14. Resolve m 6 — ?i 6 into four factors.
15. Resolve a 9 + c 9 into three factors.
16. Resolve 64 « 6 — 1 into four factors.
Other methods for factoring will he given in Chapter XXIX.
IX.— GREATEST COMMON DIVISOR.
122. A Common Divisor or Measure of two or more quan
tities is a quantity that will divide each of them without a
remainder.
Hence, any factor common to two or more quantities is a
common divisor of those quantities.
Also, when quantities are prime to each other, they have no
common measure greater than unity.
54 ALGEBRA.
123. The Greatest Common Divisor of two or more quan
tities is the greatest quantity that will divide each of them
without a remainder.
Hence, the greatest common divisor of tivo or more quanti
ties is the product of all the prime factors common to those
quantities.
By the greatest of two or more algebraic quantities, it may
he remarked, is here meant the highest, with reference to the
coefficients and exponents of the same letters.
In determining the greatest common divisor of algebraic
quantities, it is convenient to distinguish three cases.
CASE I.
124. When the quantities are monomials.
1. Find the greatest common divisor of
42 a 3 b 2 , 70 a 2 b c, and 98 a 4 b 3 d 2 .
42 a 3 b 2 =2x3x7 a a a bb
70 a 2 be =2x5x7 aa b c
98 a 4 b 3 d 2 = 2 x 7 X 7 aaaabbb del
Hence, G. C. D. = 2 X 7 a a b = 14 d 2 b, Ans. (Art. 123). •
RULE.
Resolve the quantities into their prime factors, and find the
product of all the factors common to the several quantities.
Note. Any literal factor forming a part of the greatest common divisor
will take the lowest exponent with which it occurs in either of the given
quantities.
EXAMPLES.
Find the greatest common divisors of the following :
2. a s x 2 , 7 a* x, and 3 a b 2 .
3. G c 5 d\ 3 c 3 d 5 , and 9 c* d 3 .
4. 18 m n 5 , 45 m 2 n, and 72 m 8 ri 2 .
5. 15 c 2 x, 45 c 3 x 2 , and 60 c 4 x 8 .
GREATEST COMMON DIVISOR. 55
6. 108 y 2 z\ 144 f z\ and 120 if z 5 .
7. 96 a 5 b\ 120 a 3 b 5 , and 168 a 4 b & .
8. 51 m 4 n, 85 ra 3 a, and 119 m 2 if.
9. 84 a 8 y 4 z s , 112 a; 4 v/ 5 z 6 , and 154 a 7 y 6 z\
CASE II.
125. When the quantities are polynomials which can be
readily factored by inspection.
1. Find the greatest common divisor of
5xy 3 — 15y 3 , x 2 + 4 x — 21, and mx — 3m — nx + 3n.
5xy 3 — 15y 3 = 5y 3 (x — 3)
x 2 + 4 x  21 = (x + 7) (a;  3)
mx — 3 m — nx + 3n = (m — n) (x — 3)
Hence, by Art. 123, G. C. D. = x — 3, Ans.
2. Find the greatest common divisor of
4 x 2  4 x + 1, 4 x 2  1, and 8 ar 5  1.
4a; 2 4a; + l = (2a;l) (2 a;  1)
4 a 2  1 = (2 x + 1) (2 x1)
8 a?  1 = (2 x  1) (4 a; 2 + 2 x + 1)
Hence, G. C. D. = 2 a: — 1, Ans.
The rule in this case is the same as in Case I.
EXAMPLES.
Find the greatest common divisors of the following :
3. 3 a x 2 — 2 a 2 x, a 2 x 2 — 3 a b x, and 5 a 2 x 3 + 2 ax 4 — 3 a 3 x.
4. vi 2 + 2 in n + n 2 , m 2 — n 2 , and m 3 + n 3 .
5. x* — 1, a 5 + a 3 , and a; 4 + 2 a; 2 + 1.
6. 3 a xif + 21 ay 2 , 3 c a + 21 c  3 d x  21 rZ, and a; 2 3a  70.
7. 4 x 2 — 12 a + 9, 4 a; 2  9, and 4 m 2 »a;6 m 2 n.
8. 9a 2 16, 3a;y — 4^ + 3a;2 — 4a, and 27 a; 3 — 64.
56 ALGEBRA.
9. x s — x, X s + 9 x 2 — 10 x, and x 6 — x.
10. a 3 — 8 ft 3 , 5 a x + 2 a — 10 ft x — 4 ft, and a 2 — 4 a ft + 4 ft 2 .
11. ar 2  a;  42, x 2  4 a  60, and x 2 + 12 x + 36.
12. 8 x 3 + 125, 4 aj 2  25, and 4 a 2 + 20 x + 25.
13. 3 a x G — 3 a x 5 , a x s — 9 a x 2 + 8 a x, and 2 a x h — 2 a x.
14. 12 a x  3 a + 8 c x  2 c, 64 x 3  1, and 16 as* — 8 a? + 1.
CASE III.
126. When the quantities are polynomials which cannot be
readily factored by inspection.
Let a and ft be two expressions, arranged in order of powers
of some common letter ; and let the exponent of the highest
power of that letter in ft be either equal to or less than the
exponent of the highest power of that letter in a. Suppose
that ft is contained in a, p times with a remainder c ; suppose
that c is contained in ft, q times with a remainder d ; and
suppose that d is contained in c, r times with no remainder.
The operation of division may be shown as follows :
ft) a (p
p ft
e) b (q
1 c
d) c (r
rd
We will first show that d is a common divisor of a and ft.
From the nature of subtraction, the minuend equals the sub
trahend plus the remainder ; hence,
a=pb + c, ft = q c + d, and c = rd.
Substituting r d for c in tho value of ft, we have
b = q r d + d = d (q r + 1).
GREATEST COMMON DIVISOR. 57
Substituting q r d + d for b, and r d for c in the value of a,
we have
a=p q r d + p d + r d = d (p q r + p + r).
Hence, as d is a factor of a and also of b, it is a common
divisor of a and b.
We will now show that every common divisor of a and b is
a divisor of d. Let k he any common divisor of a and b, such
that fl^ffli and b = n k. From the nature of subtraction,
the minuend minus the subtrahend equals the remainder ;
hence,
c = a — p b, and d = b — q c.
Substituting m k for a, and n k for b in the value of c, we
have
c = m k — p n k.
Substituting mk — pnk for c, and n k for b in the value of
d, we have
d — nk — q (j)ik—pnk) = nk — qmk+pqnk
= k (n — q m + p q n).
Hence, k is a factor or divisor of d.
Therefore, since every common divisor of a and b is a divisor
of d, and no expression greater (Art. 123) than d can be a
divisor of d, it follows that d is the greatest common divisor
of a and b.
1. Find the greatest common divisor of x 1 — 6 x + 8 and
4 x 3  21 x 2 + 15 x + 20.
x 2  6 x + 8) 4 a 3  21 x 2 + 15 x + 20 (4 x + 3
4 x 3  24 a; 2 + 32 x
3 x 2  17 x + 20
3 a; 2  18 a; + 24
x _ 4) ^ — 6x + 8(x2
x 2 — 4aj
. 2a; + 8
2a + 8
Hence, cc — 4 is the greatest common divisor, Ans.
58 ALGEBRA.
RULE.
Divide the greater quantity (Art. 123) by the less • and if
there is no remainder, the less quantity trill he the required
greatest common divisor.
If there is a remainder, divide the divisor by it, and continue
thus to make the preceding divisor the dividend, and the re
mainder the divisor, until a divisor is obtained which leaves no
remainder ; the last divisor will be the greatest common divisor
required.
Nota 1. If there are three or more quantities, find the greatest common
divisor of two of them ; then of this result and the third of the quantities,
and so on. The last divisor will be the greatest common divisor required.
Note 2. If a monomial factor is seen by inspection to be common to all
the terms of one of the given quantities, and not of the other, it may be re
moved, as it evidently can form no part of the greatest common divisor ;
and, similarly, we may remove from a remainder any monomial factor
which is not a common factor of the given quantities.
2. Find the greatest common divisor of
6 a x~ — 19 a x + 10 a and 6 x s — x 2 — 35 x.
In the first quantity a is a common factor of all the terms,
and is not a factor of the second quantity ; in the second quan
tity x is a common factor of all the terms, and is not a factor
of the first quantity. Hence we may remove a from each
term of the first quantity, and x from each term of the second.
6 a; 2  19 a + 10)6 cc 2  x 35(1
6a: 2  19:c + 10
18a;45
In this remainder 9 is a common factor of all the terms, and
is not a common factor of the given quantities. Hence 9 may
be removed from each term of the remainder.
2 x  5)6 x 1  19 x + 10(3 x  2
6 x 2 — 15 x
— 4 x'+ 10
— 4 a; + 10
Hence, 2 x — 5 is the greatest common divisor, Ans.
GREATEST COMMON DIVISOR. 59
Note 3. If the first term of a remainder be negative, the sign of each
term may be changed.
3. Find the greatest common divisor of 2 x 2 — 3 a; — 2 and
2a; 2 5a;3.
2a; 2  3 x  2)2 a 2  5 x 3(1
2x 2 3x2
2xl
The first term of this remainder being negative, we change
the sign of each term, giving 2 x + 1.
2 x + 1)2 a;' 2  3 x  2 (x — 2
2 x' 2 + x
— 4x — 2
— 4a; — 2
Hence, 2 x + 1 is the greatest common divisor, Ans.
Note 4. The dividend or any remainder may be multiplied by any
quantity which is not a common factor of all the terms of the divisor.
4. Find the greatest common divisor of 2 a 3 — 7 xr + 5 x — 6
and 3 a' 3 — 7 a' 2 — 7 x + 3.
To avoid a fraction as the first term of the quotient, we
multiply each term of the second quantity by 2, giving
6 X s  14 x 2  14 x + 6.
2 x s — 7 x 2 + 5 x — 6)6 x s  14 cc 2  14 x + 6 (3
6 a; 3  21 x 2 + 15 a;  18
7 a 2 29 a; + 24
To avoid a fraction as the first term of the next quotient,
we multiply each term of the new dividend by 7, giving
14 x 3  49 x 2 + 35 x  42.
7 a; 2  29 x + 24) 14 x 3  49 x 2 + 35 x  42 (2 a;
14 a; 3  58 a 2 + 48 x
9 a 2  13 x  42
60 ALGEBRA.
The first term of this remainder not heing exactly divisible
by the first term of the divisor, we multiply each term hy 7,
giving 63 x 2 — 91 x — 294.
7 x 2  29 x + 24) 63 x 2  91^294(9
63 x 2  261 x + 216
170 x  510
Dividing each term by 170, x — 3) 7 x 2 — 29 x + 24 (7 x — 8
7 x 2 21x
— 8 a; + 24
 8cc + 24
Hence, x — 3 is the greatest common divisor, Ans.
Note 5. When the two given quantities have a common monomial
factor, it may be removed from each, and the greatest common divisor of
the resulting expressions found. This result must be multiplied by the
common monomial factor to give the greatest common divisor of the given
quantities.
5. Find the greatest common divisor of 6 x 3 — x 2 — 5 x and
21 x 3  26 x 2 + 5x.
The quantities have the common monomial factor x ; remov
ing it, we find the greatest common divisor of 6 x 2 — x — 5 and
21 x 2 — 26 x + 5. We multiply the latter by 2, to avoid a frac
tion as the first term of the quotient, giving 42 a; 2 — 52 x + 10.
6 x 2  x  5) 42 x 2  52 x + 10 ( 7
42 x 2  7a;35
— 45 x + 45
Dividing by — 45, x — 1)6 x 2 — x — 5(6a: + 5
6 x 2 — 6 x
5x — 5
5x — 5
LEAST COMMON MULTIPLE. 61
Hence, x — 1 is the greatest common divisor of 6 x 2 — x — 5
and 21 x 2 — 26 x + 5. Multiplying by x, the common mo
nomial factor, we obtain x (x — 1) or x 2 — x as the required
greatest common divisor, Ans.
EXAMPLES.
Find the greatest common divisors of the following :
6. 6 x 2 — 1 x — 24 and 12 x 2 + 8 x  15.
7. 24 x 2 + 11 x  28" and 40 x 2  51 x + 14.
8. 2 x s 2 x 2  3 x + 3 and 2 x 3 2 x 2  ox + 5.
9. 6 x 2  13 x 28 and 15 x 2 + 23 a; + 4.
10. 8 x 2  22 x + 5 and 6 a; 2  23 x + 20.
11. 5 x 2 + 58 jc + 33 and 10 .x 2 + 41 x + 21.
12. x 3 + 2 x 2 + x + 2 and x i 4rx2.
13. 2a; 3 3cc 2 ;c + l and 6x s — x 2 + 3x 2.
14. a, 4  x 3 + 2 x 2 + x + 3 and x i + 2 x 3 — x — 2.
15. ft 2 5ax + 4 x 2 and a 3 — « 2 x + 3 a x 2 — 3 x 3 .
16. x A x 3 5 x 2 + 2 a + 6 and x 4 + x 3  x 2 2x 2.
17. 6 x 2 y + 4 x y 2 — 2 y 3 and 4 cc 3 + 2 x 2 ?/ — 2 a* //\
18. 2« 4 + 3a 3 x9a 2 x 2 and 6 « 3  17 a 2 ;c + 14 a ar 3x 3 .
19. 15 a 2 x 3  20 a 2 x 2  G5 a 2 x  30 a 2 and 12 6 a; 3 + 20 b x 2
— 16 b x — 16 b.
X. — LEAST COMMON MULTIPLE.
127. A Multiple of a quantity is any quantity that can be
divided by it without a remainder.
Hence, a multiple of a quantity must contain all the prime
factors of that quantity.
62 ALGEBRA.
128. A Common Multiple of two or more quantities is one
that can be divided by each of them without a remainder.
Hence, a common multiple of two or more quantities must
contain all the prime factors of each of the quantities.
129. The Least Common Multiple of two or more quanti
ties is the least quantity that can be divided by each of them
without a remainder.
Hence, the least common multiple of two or more quantities
must be the pmxluct of all their different prime factors, each
taken only the greatest number of times it is found in any one
of those quantities.
By the least quantity, is here meant the lowest with refer
ence to the exponents and coefficients of the same letters.
In determining the least common multiple of algebraic
quantities, we may distinguish three cases.
CASE I.
130. When the quantities are monomials.
1. Find the least common multiple of 36 a s x, 60 a 2 y 2 , and
84 c x s .
36 a 3 # = 2x2x3x3 a a ax
60aV=2x2x3x5 a a yy
84 c x z =2x2x3x7 x x x c
Hence, L. C. M. = 2 x f 2 X3x3x5 X? aaaxxxyyc
= 1260 a 3 x s if c, Ans. (Art. 129).
RULE.
Resolve the quantities into their prime factors; and the
product of these, taking each factor only the greatest number of
times it enters into any one of the quantities, will be the least
common multiple.
Any literal factor forming a part of the least common mul
tiple will take the highest exponent with which it occurs in
p.ither of the given quantities.
LEAST COMMON MULTIPLE. 6
o
When quantities are prime to each other, their product is
their least common multiple.
EXAMPLES.
Find the least common multiples of the following :
2. 8 a 4 c, 10 a 3 b, and 12 a 2 b 2 .
3. 5 x 3 y, 10 if z, and 15 x z 3 .
4. a 5 b 2 , 9 a 3 b\ and 12 a 2 b 3 .
5. 24 m 3 x 2 , 30 n 2 y, and 32 x y 2 .
6. 8 c 2 d 3 , 10 a e, and 42 a 2 d.
7. 36 x y 2 z 3 , 63 x 3 y z 2 , and 28 .r 2 y 3 z.
8. 40 a 2 b d 3 , 18 a c 3 d\ and 54 J 2 c d\
9. 7 m w 2 , 8 x 3 y 2 , and 84 n x y 3 .
CASE II.
131. When the quantities are polynomials which can be
readily factored by inspection.
1. Find the least common multiple of x 2 + x — 6, x 2 — 6 x + 8
and x 2 — 9.
x * + a, _ 6 = (x  2) (x + 3)
a;2_6a: + 8 = (a;2) (cc4)
x 2 9 =(x3)(x + 3)
Hence (Art. 129), L. C. M. = (x  2) (a;  3) (sc + 3) (x  4)
or, x 4  6 x 3  x 2 + 54 x  72, ^ws.
The rule is the same as in Case I.
EXAMPLES.
Find the least common multiples of the following :
2. a x 2 + a 2 x, x 2 — a 2 , and x 3 — a 3 .
3. 2 a 2 + 2 a b , 3 a b  3 & 2 , and 4 a 2 c — 4 ft 2 c.
64 ALGEBRA.
4. x 2 + x, x z — x, and x i + x.
5. 2  2 a 2 , 4  4 a, 8 + 8 a, and 12 + 12 a 2 .
6. x 2 + 5x + 4, x 2 + 2x — S, and a; 2 + 7 jc + 12.
7. x 3 — 10 a; 2 + 21 a;, and a x 2 + 5 a x — 24 a.
8. 4 ar  4 a; + 1, 4 x 2  1, and 8 a 3  1.
9. a x — a y — b x + b y, x 2 — 2 x y + y 2 , and 3 arb — 3ab 2 .
10. 9 a; 2 + 12 a; + 4, 27 x 3 + 8, and 6 a x 3 + 4 a x 2 .
11. cc 2  4 cc + 3, x 2 + a;  12, and x 2 x 20.
12. x 2 — y 2 — z 2 + 2 y z and x 2 i/ 2 +r + 2a;s.
CASE III.
132. When the quantities are polynomials which cannot be
readily factored by inspection.
Let a and b be two expressions ; let d be their greatest com
mon divisor, and m their least common multiple. Suppose
that d is contained in a, x times, and in b, y times ; then, from
the nature of the greatest common divisor, x and y are prime
to each other. Since the dividend is the product of the quo
tient and divisor, we have
a = dx and b = d y.
Then (Art. 129) the least common multiple of a and b is
d x y, or m = d x y ; but dx = a, and y = ', substituting, we
h
have m = a X ; •
d
In a similar manner we could show that m — b X ;•
d
Hence the following
RULE.
Find the greatest common divisor of the two quantities ; di
vide one of the quantities by this, and multiply the quotient by
the other quantity.
LEAST COMMON MULTIPLE. 65
Note. If there are three or more quantities, find the least common
multiple of two of them, and then of that result and the third quantity ;
and so on.
1. Find the least common multiple of 6 x 2 — 17 x + 12 and
12 a: 2  4 a 21.
6 a; 2  17 x + 12)12 a: 2  4sc21(2
12 x 2  34 x + 24
30 x  45
2 x  3 ) 6 x 2  1 7 x + 1 2 ( 3 x  4
6 x 2 — 9 x
 8 x + 12
 8 a + 12
Hence, 2 a: — 3 is the greatest common divisor of the two
quantities ; dividing the first given quantity by this, we obtain,
as a quotient, 3 x — 4 ; multiplying the second given quantity
by this quotient, we have
(3 a; 4) (12 a; 2 4 a: 21), or 36 x 3  60 x 2  47 x + 84
as the required least common multiple, Ans.
EXAMPLES.
Find the least common multiples of the following :
2. 6 x 2 + 13 x  28 and 12 x 2  31 x + 20.
3. 8 x 2 + 30 x + 7 and 12 x 2  29 x  8.
4. a 3 + a 2_ 8 a _ 6 and 2 a 3  5 a 2  2 a + 2.
5. 2 x 3 + x 2  x + 3 and 2 .t 3 + 5 x 2  x  6.
6. (t 3 2« 2 H2ffii 2  6 3 and a 3 + a 2 6  a b 2  b 3 .
7. x* + 2 x 3 + 2 x 2 + x and a « 3 — 2 a x — a.
8. 2x 4 llx 3 +3a; 2 + 10a; and 3a; 4  14z 3  6ar+ 5oj.
66 ALGEBRA.
XL — FRACTIONS.
133. A Fraction is an expression indicating a certain
number of the equal parts into which a unit has been divided.
The denominator of a fraction shows into how many parts
the unit has been divided, and the numerator how many parts
are taken.
134. A fraction is expressed by writing the numerator
above, and the denominator below, a horizontal line. Thus,
 is a fraction, signifying that the unit has been divided into
b equal parts, and that a parts are taken.
The numerator and denominator are called the terms of a
fraction.
Every integer may be considered as a fraction whose denomi
a
nator is unity ; thus, a = r .
135. An Entire Quantity is one which has no fractional
part ; as, ab, or a — b.
136. A Mixed Quantity is one having both entire and
b a
fractional parts ; as, a , or c +
x + y
137. If the numerator of a fraction be multiplied, or the
de nominator divided, by any quantity, the fraction is multi
plied by that quantity.
1. Let y denote any fraction ; multiplying its numerator by
c, we have — . Now, in  and — the unit is divided into b
b b b
equal parts, and a and a c parts, respectively, arc taken. Since
FRACTIONS. 67
c times as many parts are taken in — as in  , it follows that
a c . .,. a
—  is c times .
b o
2. Let — denote any fraction ; dividing its denominator
a a a
by c, we have . Now, in — and , the same number of
b be a b
parts is taken ; but, since in — tbe unit is divided into
a . hc
b c equal parts, and in  into only b equal parts, it follows that
• a . . , i , • a tt a
each part m  is c times as large as each part m — . Hence, 
is c times r— .
be
138. If the numerator of a fraction be divided, or the de
nominator multiplied, by any quantity, the fraction is divided
by that quantity.
1. Let — denote any fraction ; dividing its numerator by c,
we have. Now, in Art. 137, 1, we showed that — was c
a o a . ac . . b
times . Hence,  is — divided by c.
b b . b
2. Let  denote any fraction : multiplying its denominator
b
a a
by c, we have — . Now, in Art. 137, 2, we showed that  was
a bc a a . . h
e times — . Hence, — is  divided by c.
b c b c b
139. If the terms of a fraction be both multiplied, or both
divided by the same quantity, the value of the fraction is not
altered.
For, multiplying the numerator by any quantity, multiplies
the fraction by that quantity ; and multiplying the denomi
nator by the same quantity, divides the fraction by that
quantity. And, by Art. 44, Ax. 6, if any quantity be both
multiplied and divided by the same quantity, its value is not
altered.
68 ALGEBRA.
Similarly, we may show that if Loth terms are divided by
the same quantity, the value of the fraction is not altered.
140. We may now show the propriety of the use of the
fractional form to indicate division, as explained in Art. 16 ;
ft
that is, we shall show that  represents the quotient of a di
vided by b.
For, let x denote the quotient of a divided by b.
Then, since the quotient, multiplied by the divisor, gives
tbe dividend, we have b x = a.
But, by Art. 137, bXj=a.
Therefore, x =  .
b
141. A fraction is positive when its numerator and de
nominator have the same sign, and negative when they have
different signs.
For, a fraction represents the quotient of its numerator
divided by its denominator ; consequently its proper sign can
be determined as in division (Art. 91).
142. The Sign of a fraction is the sign prefixed to its
dividing line, and indicates whether the fraction is to be
added or subtracted.
Thus, in x \ — — the sign + denotes that the fraction j— ,
although essentially negative (Art. 91), is to be added to x.
The sign written before the dividing line of a fraction is
termed the apparent sign of the fraction ; and that de] tending
upon the value of the fraction itself is termed the real sign.
Thus, in \ — — , the apparent sign is + , but the real sign
is — .
Where no signs are prefixed, plus is understood.
ah
— ah ah
ah
h
b b
b
ah
a b — ah
— ah
— _ i
b
b b
b
FRACTIONS. 69
143. From the principles of Arts. 140 and 141 we obtain,
+ a;
— a.
— o
From which it appears that,
Of the three signs prefixed to the numerator, denominator,
and dividing line of a fraction, any two may he changed with
out altering the value of the fraction ; hut if any one, or all
three are changed, the value of the fraction is changed from
+ to — , or from — to + .
144. If either the numerator or denominator of the frac
tion is a polynomial, we mean by its sign the sign of the entire
expression, as distinguished from the sign of any one of its
individual terms • and care must be taken, pn changing signs
in any such case, to change the sign before each term.
„,, a — h—a + b b — a
Thus ' c^rc^d> ov c—}v
a—h—a+b h — a
also, j = ., or .
c — d —c + d d — c
145. From Art. 141 we have
abed _(a)h ( c) (—d) __ a ( b) ( c) d
efffh (~e)fgh e(f)g(h)
, etc. ;
abed _ (— a) bc{— d) _ a (— h) (— c) d
~ Jff h  ( r e)fgY ~ e (/) ( g) ( h)' :
From which it appears that,
If the terms of a fraction are composed of any number of
factors, any even number of factors may have their signs
changed without altering the value of the fraction • but if any
70 ALGEBRA.
odd number of factors have their signs changed, the value of
the fraction is changed from + to — , or from — to +.
a — b a — b b — a
Thus,
(x — y)(x — z)~ (y — x) (z — x)~ {y^ x) {x — z)
b — a 1 n . b — a
but does not equal
(x — y)(z — x) (y — x){z — x)
REDUCTION OF FRACTIONS.
146. Reduction of Fractions is the process of changing
their forms without altering their values.
TO REDUCE A FRACTION TO ITS SIMPLEST FORM.
147. A fraction is in its simplest form, when its terms are
prime to each other.
CASE I.
148. When the numerator and denominator can be readily
factored by inspection.
Since dividing hoth numerator and denominator by the
same quantity, or cancelling equal factors in each, does not
alter the value of the fraction (Art. 139), we have the fol
lowing
RULE.
Resolve both numerator and denominator into their prime
factors, and cancel all that are common to both.
EXAMPLES.
, ^, , 18 a 8 b 2 o . . , ,
1. ixeduce — — ^— 2 — ™ ^ s simplest form.
18 a*b 2 c __ 2 . 3 . 3 . a . a . a . b . h . c 2ac
45 a 2 b' 2 x ~ 5 . 3 . 3 . a . a . b . b . x 5 x ' *
x 2 + 2 x — 15
2. Eeduce — s — — to its simplest form.
x — 2x — 3
g; 2 + 2x 15 _ (x + 5) (x  3) _ ,r + 5
x 2  2 x  3 : " (x + 1) (x  3) ~ x~+l'
FRACTIONS. 71
_ _. , b c, — a r — b d + a d . . .
3. Eeduce : ; — to its simplest form.
a in — b in — an + n
be — ac — bd + ad (b — a) (c — d)
am — b m — an+ bn (a — b) (m — n)
= (Art. 89) ("f)(<*«0 = fLll t Ans .
(it — 0) (m — n) m — n
Note. If all the factors of the numerator be removed by cancellation,
the number 1 (being a factor of all algebraic expressions) remains to form a
numerator.
If all the factors of the denominator be removed, the result will be au
entire quantity ; this being a case of exact division.
Reduce the following to their simplest forms :
4. SfUU. 13.
c K.9U !IV IV  
oo mr n 6
65x 2 y 3 z*
2(> x* y° z~
„ 54 a 3 b 5 c 2
72 a 2 b 2 c
Wmxif ,«
o. — — • 1 1 .
to m x y
110 e 3 x 2 y
9  22c 2 x 2 ■ 1S '
1A 2a 2 cd+2abcd 1Q
1U  a~^ Tr — 7 ' iy>
b <r x y + b ab,x y
ii. 3»'e« 4 y go.
6x 2 y 2 — 12 xy 3 ac + ad — b c — b d
19 x 2 — 2x — lh' 2mx + 3my — 2n 2 x — 3n 2 y
x 2 + 10 x + 21 ' 2 m 2 x + 3 m 2 y—2nx—3n y
m 2 — 10m
+ 16
m 2 + m 
72 *
4 c 2 20c
' + 25
254
c 2 '
4 a —
9 a n 2
9bn 2 12bn + 4:b
8 x 3 + y 3
4 x 2 — y 2 '
27 y 3  125
25
30 7/ + 9y 2 '
6 a; 2
y — 2 x 3 y
x 2 
8x + 15 "
4 — x 2
X 3 
 9 x 2 + 14 x '
a e
— b c — ad + bd
72 ALGEBRA.
CASE II.
149. When the numerator and denominator cannot be
readily factored by inspection.
Since the greatest common divisor of two quantities con
tains all the prime factors common to both, we have the fol
lowing
KULE.
Divide both numerator and denominator by their greatest
common divisor.
EXAMPLES.
1. Reduce —  — 5 — — to its simplest form.
6 cr — a — 12
By the rule of Art. 126, we find the greatest common divisor
of the numerator and denominator to he 2.a — 3. Dividing
the numerator by this, the quotient is a — 1. Dividing the
denominator, the quotient is 3 a + 4. Therefore, the simplest
form of the fraction is , Ans.
3 a + 4'
Reduce the following to their simplest forms :
6a: 2 + a:35 „ 6 a 3  19 x 2 + 7 x + 12
o «" " — " — "J g
' '2u 2 7a + 6'
. 2 m 2 — 5 m + 3 Q
' 12 m 3  28 m + 15 '
x a + x *Sx2
O. = ; s ~ 7Z • 1U.
6 ' 2x 3 + 5x 2 2:c + 3"
8x 2 +
22 a: + 5'
10 a 2 
a21
2u 2 
7 a + 6'
2 m 2
— 5 m +
3
12 m 2 
 28 m +
15 '
x s + x 2 — 3 x 
2
x 3 — 4
x 2 + 2x
43"
6 a 3 —
7 x 2 + 5 a
•2
6a; 3 25a: 2 + 17a +
20
4a: 3 + 14ar+12a +
5
4a; 3 10a; 2 12a'
rr'
1
12 a 2 + 16 a  3
10 a 2 + a  21 '
x 3 — 4 x 2 + 4 a; — 1
a 3  2 x 2 + 4 aj  3 "
6 x 3 — x 2 — 7 x — 2
FRACTIONS. 73
TO REDUCE A FRACTION TO AN ENTIRE OR MIXED QUANTITY.
150. Since a fraction is an expression of division (Art.
140), we have the following
RULE.
*
Divide the numerator by the denominator, and the quotient
will be the entire or mixed quantity required.
EXAMPLES.
ax — a 2 x 2
1. Reduce — to an entire quantity.
(ax — a 2 x 2 ) iax = l — ax, Ans.
q% A3 /j.3
2. Reduce to a mixed quantity.
b 3
a — x)a 3 — x 3 — b 3 (a 2 + a x + x 2 , Ans.
a — x
a 3 — a 2 x
a 2 x —
x 3 —
b 3
9
a x —
a x 2
ax 2
— X 3
—
b z
ax 2
— x 3
b 3
Reduce the following to entire or mixed quantities :
ab — a 2 o~ 2 ■ k
o.
4.
5.
6.
7.
2ab
b '
X 3
+ y 3
X
+ y
2:
K 2 3x4:
5 x
X 3
2 i rr
— x' + 7x —
6
3 x
a 2
— 3ab + 4b 2
8.
x — 3
9.
X s — 1
XV
10.
4,x 2 2x + 5
2x 2 x + \ '
11
/Y»<* ___ ryi"— /y» i i _ */
*As *Aj *as *J
X 2 + X — 1
12
2cc 3 3.x 2 +4r2
2 x 2 — 3 x + 3
74 ALGEBRA.
TO REDUCE A MIXED QUANTITY TO A FRACTIONAL FORM.
151. This is the converse of Art. 150; hence we may
proceed by the following
RULE.
Multiply the entire part by the denominator of the fraction ;
add the numerator to the product when the sign of the fraction
is + , and subtract it when the sign is — / writing the result
over the denominator.
EXAMPLES.
a 2 £2 5
1. Eecluce a + b — : — to a fractional form.
a — b
By the rule,
a 2 _ b 2 _ 5 ( a + &) (gb) Q 2  b°  5)
a — b a — b
a — b a — b
Note. It will be found convenient to enclose the numerator in a pa
renthesis, when the sign before the fraction is — .
Reduce the following to fractional forms:
4 „ 3a 2 2Z> 2
2. x + l + . 7.2a ^
x — 6 a a
3. a + — 8. a 2 +ab + b 2 — 7 •
n b — a
4. 7a;  4 " 2 + 5a  9. 3z2 3
~8 2xl
i x + 1 in / a * + b *
5. * + i + __. 10. a &_^.
a +■ « a; — ^s
FRACTIONS. 75
TO REDUCE FRACTIONS TO A COMMON DENOMINATOR.
t co 1 x> i 5cd 3mx j 3 n y
152. 1. Keduce  — — , , and — ^ to a common
3 crb 2 ab 2 ka 6 b
denominator.
Since multiplying each term of a fraction by the same quan
tity does not alter the value of the fraction (Art. 139), we
may multiply each term of the first fraction by 4 a b, giving
20 a b c d , « , 1 ■, . „ . . 18 a 2 m x
; each term of the second by b a , giving ;
12 a? 6 2 ' J ' & ° 12« :i 6 2
and each term of the third by 3 b, giving L. .
12 a J b"
It will be observed that the common denominator is the
least common multiple of the given denominators, which is
also called the least common denominator ; and that each term
of either fraction is multiplied by a quantity which is obtained
by dividing the least common denominator by its own denomi
nator. Kence the following
RULE.
Find the least common multiple of the given denominators.
Divide this by each denominator, separately, and multiply the
corresponding numerators by the quotients ; writing the results
over the common denominator.
Before applying the rule, each fraction should be in its sim
plest form ; entire and mixed quantities should be changed to
a fractional form (Arts. 134 and 151).
Note. The common denominator may be any common multiple of the
given denominators. The product of all the denominators is evidently s
common multiple, and the rule is sometimes given as follows : "Multiply
'each numerator by all the denominators except its own, and write the
results over the product of all the denominators."
an a x x it
2. Reduce  — — , — — , and . . ■ N , to a common de
1 — x (1 — x) 2 (1 — xy
nominator.
76 ALGEBRA.
The least common multiple of the given denominators is
(1 — be) 3 . Dividing this by the first denominator, the quotient
is (1 — a:) 2 ; dividing it by the second denominator, the quo
tient is (1 — x) ; and dividing it by the third denominator, the
quotient is 1. Multiplying the corresponding numerators by
these quotients, we obtain a y (1 — x) 2 , a ar (1 — x), and x i/ 3
as the new numerators. Hence the results are
a y (1 — a?) 2 ax 2 (1 — x) x y 3
(ixy > {ixy > and (i  xy ' Ans 
EXAMPLES.
Reduce the following fractions to a common denominator :
a 3ab 2ac ,56c _ 4c — 1 3b — 2 1 5a
3  ^o— > n— > and "To 6  o „ , > K „ . , and
8 ' 9 ' 12 ' 3ab ' 5ac ' 6b c
. x 2 y xy z 7 y z 2 „ 2 3 4
^** ~t , * i t~^ i and ^"tt — . /. —  — — , — , and — .
10 ' 15 ' 30 a 3 a; 2 ' a a: 3 ' at
a 2 x
3y z Axz 5a;?/ 5 az 3bx ley — m
b ' Yx'Yy' and TT " 8  6^' 87i' and 10* s 2
9. , — — , and —
a — b ' a + b ' a 2 + 6 2 °
10 # + 3 a? + 1 a; + 2
a; 2  3 a: + 2' a; 2  5x + 6' x 2  4a; + 3'
2a 3b 4c
cr + a — 6 ' a 2 + 5a + 6' an< " a 2 — 4'
12. T , — — T , and j — T .
a; — 1 ar— 1 x 3 — 1
n a & m — n a + b
a m — b m + a n — b n' 2 a 2 — 2 a b ' 3 b m + 3 b n
14. Reduce ; r^—  , — ;—  , and
(ab) (ac) ' (b — a) (be)' (ca)(cb)
to u common denominator.
FRACTIONS. 77
The fractions may be written (Art 145) as follows :
, and
(a  6) (a  c) ' (a b)(b e) ' (a  c) (6  c)
The least common denominator is now (a — b) (a — e)
(b — c). Applying the rule, we have the results,
Qa)(bc) (hl)(ac) ^ ^
(a b) (a c) (b — c)' (a b) (a  c) (b  c)
(lc)(ab)
(a b){a c) (b  c)
Kecluce to a common denominator :
,_ 3 2 .a — 2
3
, Ans.
a
—
V
a +
V
1
2
X
1
+
X
c
>
x —
+ d
1'
16. z , 7, and
x'
1 — x . b — a
17. . j^. pr, p. sT? ~ , and
(a + b)(ab)' (ba)(cd)' (dc)(a + b)
153. A fraction may he reduced to an equivalent one hav
ing a given denominator, by dividing the given denominator
by the denominator of the fraction, and multiplying both terms
by the quotient.
1. Eeduce — ry to an equivalent fraction having
ar — a b + b"
a 3 + b 3 for its denominator.
(a* + b 3 ) j (a 2 ab + b 2 ) = a + b;
multiplying both terms by a + b,
a — b (a  b) (a + b) a 2 b 2
a*ab + b* ~ (a 2 ab + b 2 ) (a + b) ~ a 3 + b 3, **'
78 algesra.
EXAMPLES.
2. Reduce to a fraction with a 2 — b 2 for its denom
a + b
inator.
x \ 1
3. Reduce  to a fraction with x 2 + 5 x — 24 for its
x — 3
denominator.
. r, , 3 m + 2
4. Reduce to a fraction with G m? — 19 m + 10
for its denominator.
4
5. Reduce to a fraction with a 3 — b 3 for its denom
a —
inator.
6. Reduce 1 + x to a fraction with 1 — x for its denomi
nator.
ADDITION AND SUBTRACTION OF FRACTIONS.
154. 1. Let it he required to add  to .
c c
In  and  , the unit is divided into c equal parts, and a
and b parts, respectively, are taken, or in all a + b parts ; that
a + 6
is . Ihus,
c
a b a + b
 +  =— I— .
c c c
2. Let it he required to subtract  from  .
c c
The result must he such a quantity as when added to 7 will
produce ; that quantity is evidently ■ • (Art. 154, 1).
,„, a b a — b
1 hus, = .
c c c
Hence the following
FRACTIONS. 79
RULE.
To add or subtract fractions, reduce them, if necessary, to a,
common denominator. Add or subtract the numerators, and
write the result over the common denominator.
The final result should be reduced to its simplest form,
wherever such reduction is possible.
3b  a b + a . 1  4 o 2
1. Add —  , n 7 , and —  — — .
3a ' 2b iab
The least common multiple of the denominators is 12 a b.
Then, by the rule of Art. 152,
3b a b + a 1  4 b* 12 b 2  4 ab 6 a b + 6 a 2
+ "ITT" + , , = " ToT  +
+
3a 2 b ±ab 12 a b 12 a b
3  12 lr 12 fr 2 4 a 6 + 6 a fr + 6 a 2 + 312 b 2
12 a b 12 a b
6a 2 +2ab + S
12 a b
, Ans.
n n i 4 flj — 1 . 6 « — 2
«. subtract — ^ from — .
2 x 6 a
The least common denominator is 6 ax.
6a —2 4 a; — 1 12 a a; — 4 a: 12 ax — S a
Then,
3a 2a? Gaa; 6 a x
12 a .r — 4 x — (12 ax — 3 a) 12 a a; — 4 a; — 12 a x + 3 a
6 a x 6 ax
3 a — 4 a;
6 a a;
, Ans.
Note. When a fraction whose numerator is not a monomial is preceded
by a  sign, it will be found convenient to enclose its numerator in a pa
renthesis before combining with the other numerators. If this is not done,
care must be taken to change the sign of each term in the numerator before
combining.
80 ALGEBRA.
4« 2 l 3«i 2 2 5« 2 c 2 +3
3. Simplify
2 a c 3 b 2 c 5 a
3
The least common denominator is 30 a b 2 c 3 .
4a 2 l _ 3 a b 2  2 _ 5a 2 c 2 +3
2 a c 3 b 2 c 5 a c 3
60 a 2 b 2 c 2  15 b' 2 c 2 _ 30 a 2 b 2 c 2  20 a c 2 30 a 2 b 2 c 2 + 18 If
30 a b 2 c s 30 d b 2 c 3 30 a 6 a c 3
_ 60 a 2 b 2 c 2  15 b 2 c 2  (30 a 2 b 2 c 2 20a c 2 )  (30 a 2 b 2 c 2 +18 b ? )
30 a 6 2 c 3
_ 60 a 2 & 2 c 2  15 6 2 c 2  30 a 2 b 2 c 2 + 20 a c 2  30 a 2 b 2 c 2  18 b 2
30 a 6' 2 c 3
20 a c 2  15 6 2 c 2  18 b 2
30 a b 2 c 3 > AnS '
EXAMPLES.
Simplify the following :
. 2x — 5 3 a + 11 _ a — b 2a + b b — 3a
4. \ . 9. 1 1 .
12 18 4^6 8
3 1 a 2 + 1 6 a 3 + 1 62
* 5 a J 2 + 2a 2 i' '3 a 2 12 a 3 + ITT'
2 a + 3 3 a + 5 2a;l 2x + 3 6cc + l
6 8 ' "12 ~H~ ^2(P"
ra — 2 2 — 3?»,?i 2 m + 2 m + 2 m + 3
' 2m » 3 m 2 n 3 ' ' ~J~ ~U~ ~2lT'
b — 4a a + 5b 10 2 2x — 1 3.r 2 +l
o. — — 1 — —  — . lo.
24: a ^ 30b ' '3 6x 9 a; 2 '
,. a — 2 3cc + l 6 a: 5 3
14. ■ \ '
2 + 3 4 5
3» + l 2&1 4<?l 6^+1
12 a ~~8lT" + 16 c 24^'
16. Simplify
FRACTIONS. 81
2x + l 3x — 1 11
2 x (x — 1) 3 a; (a? + 1) 4 (a 2  1)
The least common denominator is 12 x (x 2 — 1).
2x + l 3x — 1 11
Then,
2 x (x  1) 3 x (x + 1) 4 (a; 2  1)
6 (a + 1) (2 x + 1) 4 (as  1) (3 x — 1) 33 a;
12x(x 2 l) 12x(x 2 l) 12x(x 2 l)
12 x 2 + 18 x + 6 12 x  16 x + 4 33 x
12 x (x 2  1) 12 x (x  1) 12 x (x 2  1)
_ 12 x 2 + 18 x + 6  (12 x 3  16 x + 4)  33 x
12 x (x 2  1)
x + 2
12 x (x 2  1)
Aiis.
Simplify the following :
»;_* + *_. »i±» + s=»
x + 2 3 — x a — 6 a + 6
18. i L_. 20.^^.
x + 7 x + 8 1 — xl + x
a /> 2 a 5
«1. — —7 H j H — o To •
a + a — a' — b"
1 1 2x
22. +
x + y x — // x' + y
1 x 3
23. 5 ^ r +
24.
X — 1 x 2 — 1 X 3 — 1
2 x — 6 x + 2 x + 1
x 2 +3x + 2 x 2 2x — 3 x 1 — x— 6'
x x 2 x
25. Simplify — — r + — h 5
x + 1 1 — x x' 2 — 1
82 ALGEBRA.
The expression may be written (Art. 143) as follows :
X X Li X
+
X + 1 x — 1 X* — 1
The least common denominator is ar — 1.
ihen > ZTT^TZ. T + ~2 7 = 31 i— Z72 T + Z2
x + 1 x — 1 ar — 1 a;' 2 — 1 x' —1 x 2 — 1
or — a: — (.t 2 + cc) + 2 a;
x 1 — 1
Simplify the following :
x*
= 0, Ans. (Art. 102).
3 4
26. 2+*. 28. ^_+ —
a — & b — a 3 x — a 2 # 2 — 9
_,_. o rt p x o a — J. .. x x x
27. 1 — . 29. 1
3a+3^22a 1+x 1x
1 1 1
30. T^fTi z +
x
31.
(a — b){b — c) (b — a) (a — c) (c  a) (c — b)
2 3 1
(a;  2) (a;  3) (3 *) (4 a;) (a;  4) (2 a;) '
MULTIPLICATION OF FRACTIONS.
155. We showed, in Art. 137, that a fraction could be
multiplied by an integer either by multiplying its numerator
or by dividing its denominator by that integer. We will now
show how to multiply one fraction by another.
Let it be required to multiply  by  .
Let  = x, and  = y ;
where x and y may be either integral or fractional.
FRACTIONS. 83
Since the dividend equals the product of the divisor and
quotient.
a = b x, and c = d y.
Therefore, hy Art. 44, Ax. 3, a c = b d x y.
Regarding a c as the dividend, b d as the divisor, and x y as
the quotient, we have
a c
xy = Vd
Therefore, putting for x and y their values,
a c a c
~b X d^bd'
Hence the following
RULE.
Multiply the numerators together for the numerator of the
resulting fraction, and the denominators for its denominator.
Mixed quantities should be reduced to a fractional form
before applying the rule.
When there are common factors in the numerators and
denominators, they should be cancelled before performing the
multiplication.
EXAMPLES.
 ,, 1A . ■ , 6x 2 y 10 a 2 y . 3b*x*
1. Multiply together 5 — ^ , == —  , and . ~ .
1 J & 5 a 3 b 2 ' 9 b x ' 4 a y 2
6x 2 y 10 a 2 y 3 J 4 x s 6 x 10 X 3 a 2 b* x 5 y 2 b x* .
5a 3 b 2 9bx A lay 2 ' 5 X 9 X 4 « 4 b 3 x y 2 ' ' a 2 ' "
Multiply together the following :
a 2 5 c . a 3 b 2 . 3 abx 2 5 x y 2
• — 2 and —3 — j • 4  5 — 2 ancl 5 — r 
ra ?r m d n d bay 2, ab x
3 a 3 x ,4fflJ _ m ?/ n , a x
3 ' ^li and KTZ, ■ 5  7^1 and
7 A 4 5Am 4 « x m y n
84 ALGEBEA.
e 2a 6c ,5b 3ab 2 3ac 2 ,Sa<P
o6 5a be 4cd 2&d 9&c
„ 8 a 15 y 2 ,3 s; 4 _ 3 m* 2 n* . 11 z 2
'• fT~3> i7 — 5 ? and in q • "• ,, o > o — .and; — 5.
9 if lb z s ' 10 cc 3 2a; 2 '3??i' 4w 2
10. Multiply together
ar 2 _ 2cc x 2 9 . a 2 + a;
, and
a;2_2a;3' a; 2 *' cc 2 +a:.6'
r2x x 2 9 cc 2 + ^
Xs X
ce 2 — 2 cc — 3 x — x x 2 + x — 6
x(x2)(x + 3) (a  3) x (x + 1) x
(x3)(x + l)x(x 1) (a + 3) (x 2) ~ x~^l '
Multiply together the following :
1t 3.r 2 cc , 10
11. = and
5 2cc 2 4a;'
,. 4 a; + 2 5cc
1*. — ^ and
2 x + 1 '
1Q a 2 2ab + b 2 . b
lo. ; and
a + b ax — bx
., . a — b . a 2 — b 2
14. =— — and
a 2 + a b a 2 — a b
, _ 1 — x 2 1 — y 2 ,
10. q , s , and
1 + 1/ ' X + X 2 ' 1 — X
. x 2 W . x 2 25
lb. „ — and 5  — .
X s + ox r4a;
a 3 — a 2 + a x s — 8
17. r and — 5 •
x 2 + 2 x + 4 a s + 1
,_ a: 2 + 5.r+G , x 2 — Ix
18> x^I^^Yl and ^4'
FRACTIONS. 85
4 5 x
19. 1 H — and —
x x 1 x 2 — 8x + 7
20. 41 and
«X/ fcC
2 5cc + 6"
a,a _ 3 x + 2 a 2  7 a; + 12 a 3  5 x 2
2L aj»_8aj + 15' * 2 5x + 4' and a* 4 '
22. ^ 2— j Lj ^r> and 1 + ^—
a; — x y + y x~ + x y + y x — y
a* _ U 1  c 2 + 2 5 c a 2pc 2 2bc
i6  a  + cb 2 + 2ac a 2 + c 2 b 2 2ac'
OA a + b a — b 4 b 2 a + b
a — b a + b a — b Jo
_„ 2x\y . y x 2 .. x 2 — y 2
25. — — — 1 ^ = r and = — V
x + y y — cc x* — y* x' + y
DIVISION OF FRACTIONS.
156. We showed, in Art. 138, that a fraction could be
divided by an integer either by dividing its numerator or by
multiplying its denominator by that integer. We will now
show how to divide one fraction by another.
Let it be required to divide  by  .
ft n
Let x denote the quotient of $—.
b a
Then, since the quotient multiplied by the divisor gives the
dividend, we have
c a xe a
XX d = b ] ° r ' ~d = b'
86 ALGEBRA.
Multiplying each of these equals by  (Art. 44, Ax. 3),
G
Therefore,
a d
x =
be
a
c ad
d be'
Hence the following
RULE.
Invert the divisor, and proceed as in multiplication.
Mixed quantities should he reduced to a fractional form,
before applying the rule.
EXAMPLES.
6 aH , 9ab 3
1. Divide r . , by  T7 r — = — r.
5 x s y* J 10 x 2 y 5
6a 2 b 9ab 5 6a 2 b 10 x 2 y h Any .
: ^ y l_ — "L Ant
5a: 3 ?/ 4 * 10.x 2 ?/ 5 5x s y** 9ab s 3b 2 x'
x 2 — 9 x + 3
2. Divide —zrs — by — p — .
x 2 — 9 x + S _ (x + 3)(x 3) 5 _ x — S
~TE~ '' ~5~~~~ ~16~ ~ X x~+3~~ 3~ '
Divide the following:
7 ??i 2 3n 2 x 2 — y 2 x 2 +xy
6  ~2~ y "13" ' ^2^ + y 2 y ary *
. 7« 3 i . 14 a b* _ n Sy 2 5 V
1 J wr ?r o ra w x z — y x — y
5.
18 7?? a: 8 1 6 m 2 a: 4
25 wy 2 } 5« 2 2/ 6 '
1.1
*■ a 2 +2a15 } a 2 2a3'
6.
1 4t x 2 x
4"^ by 12 + 3
in a; 3 — 4 a* , x 2 —3x\2
' a; 2 + 5x + 6 b} « a +2aj3'
FRACTIONS. 87
COMPLEX FRACTIONS.
157. A Complex Fraction is one having a fraction in its
numerator, or denominator, or both. It may be regarded as a
case in division, since its numerator answers to the dividend,
and its denominator to tbe divisor.
However, since multiplying a fraction by any multiple of
its denominator must cancel that denominator, to simplify a
complex fraction, we may multiphj both of its terms by the
least common multiple of their denominators.
EXAMPLES.
a
1. Reduce  to its simplest form.
FIRST METHOD.
Proceeding as in division,
a
c a b a b
=^Xj=—t, Ans.
a c a ca
b
SECOND METHOD.
Multiplying both terms by the least common multiple of
their denominators,
a a
Xbc
e c a o .
= = —T) Ans.
da, c a
» b xbe
a a
2. Reduce — ^— — to its simplest form.
b a
a — b a + b
88 ALGEBEA.
The least common multiple of the denominators is a 2 — b 2 .
Multiplying each term by a' 2 — lr, we have
a (a \b) — a (a — b) cr + ab — a 2 + ab 2 a b
Ans.
b (a + b) + a (a — b) a b + b~ + a 2 — a b a 1 + b 2 '
3. Reduce  — to its simplest form.
x
1 1 X+ 1 33 + 1
, Ans.
1 1 _, a* a: + 1 + a; 2cc + 1
.1 a; + 1
1 +
Reduce the following to their simplest forms :
4. L. 8. !. 12
5.
b
a\ —
c
m
x
n
6.
n
» g
cc
y^ + 9
7. 5=4. 11. 15
1 *
1 +
X
1
x 2 +
X
1 '
1 +
X
a
b
b
a
1
1*
b
a
1
x 2 +
2
X
—
2/
w + ?i 1 « + ft b + .b 2 '
n 12
x — 7 \
9. j. 13. ,
a_q 18
a; + 3
X
10.
14.
1
1 — X
1
1 + £C
1
= H
1
=
31
1+  c + 1
3 — x
SIMPLE EQUATIONS. 89
a 2 + b 2 a 2 — lr in — n m 3 — n s
, n a 2 — b 2 a 2 + b 2 10 m + n in 3 + n 3
16. ; =— • !"• ~? '?•
a + b a — b m + n rnr + nr
+
a — b a + b m — n mr — n
.2
i7. x + y y , i9.
•> «2
x + 2 y x 2 "
a;
v/ a; + y a + x
158. In Art. 42, we defined the reciprocal of a quantity
as being 1 divided *by that quantity. Therefore the reciprocal
in it
of — = — = — ; or, the reciprocal of a fraction is the frac
n in m
ii
tion inverted.
XII. — SIMPLE EQUATIONS.
159. An Equation is an expression of equality between
two quantities. Thus,
x + 4 = 16
is an equation, expressing the equality of the quantities x + 4
and 16.
160. The First Member of an equation is the quantity on
the left of the sign of equality. The Second Member is the
quantity on the right of that sign. Thus, in the equation
x + 4 = 16, x + 4 is the first member, and 16 is the second
member.
The sides of an equation are its two members.
161. An Identical Equation is one in which the two mem
bers are equal, whatever values are given to the letters in
volved, if the same value be given to' the same letter in every
part of the equation ; as,
90 ALGEBRA.
2a + 2bc = 2(d + bc).
162. Equations usually consist of known and unknown
quantities. Unknown quantities are generally represented by
the last letters of the alphabet, x, y, z; but any letter may
stand for an unknown quantity. Known quantities are repre
sented by numbers, or by any except the last letters of the
alphabet.
163. A Numerical Equation is one in which all the known
quantities are represented by numbers ; as,
2 x — 11 = x — 5.
A Literal Equation is one in which some or all the known
quantities are expressed by letters j as,
2x + a = bx 2 — 10.
164. The Degree of an equation containing but one un
known quantity is denoted by the highest power of that
unknown quantitj' in the equation. Thus,
> are equations of the first degree.
and c x = a' 2 + b d ' )
3 x 2 — 2 x = 65 is an equation of the second degree.
In like manner we have equations of the third degree, fourth
degree, and so on.
When an equation contains more than one unknown quan
tity, its degree is determined by the greatest sum of the
exponents of the unknown quantities in any term. Thus,
x + x y = 25 is an equation of the second degree.
a; 2 — y 2 z = a b 3 is an equation of the third degree.
Note. These definitions of degree require that the equation shall not
contain unknown quantities in the denominators of fractions, or under
radical signs, or affected with fractional or negative exponents.
SIMPLE EQUATIONS. 91
165. A Simple Equation is an equation of the first degree.
166. The Root of an equation containing' but one unknown
quantity is the value of that unknown quantity; or it is tin
value which, being put in place of the unknown quantity,
makes the equation identical. Thus, in the equation
3x — 7 = x + 9,
if 8 is put in place of x, the equation becomes
24  7 = 8'+ 9,
which is identical; hence the root of the equation is 8.
Note. An equation may have more than one root. For example, in
the equation
x 2 = 7 J! 12,
if 3 is put in place of x, the equation becomes 9 = 2112; and if 4 is put
in place of x, it becomes 16 = 28 — 12. Each of these results being iden
tical, it follows that either 3 or 4 is a root of the equation.
167 It will be shown hereafter that a simple equation has
but one root; an equation of the second degree, two mots;
and, in general, that the degree of the equation and the num
ber of its roots correspond.
168. The solution of an equation is the process of finding
its roots. A root is verified, or the equation satisfied, when,
the root being substituted for its symbol, the equation becomes
identical.
TRANSFORMATION OF EQUATIONS.
169. To Transform an equation is to change its form with
out destroying the equality.
170. The operations required in the transformation are
based upon the general principle deduced directly from the
axioms (Art. 44) :
92 ALGEBRA.
If the same operations are performed upon equal quantities,
the results will be equal.
Hence,
Both members of an equation may be increased, diminished,
multiplied, or divided by the same quantity, without destroy
ing the equality.
TRANSPOSITION.
171. To Transpose a term of an equation is to change
it from one member to the other without destroying the
equality.
172. Consider the equation x — a = &.
Adding a to each member (Art. 170), we have
x — a + a = b + a
or, x = b + a,
where — a has been transposed to the second member by
changing its sign.
173. Again, consider the equation x + a = b.
Subtracting a from each member (Art. 170), we have
x + a — a = b — a
or, x — b — a,
where a has been transposed to the second member by chang
ing its sign.
174. Hence the following
EULE.
Any term may he indisposed from one member of an equa
tion t<> the other, provided its sign be changed.
Also, if the same term appear in both members of an equa
tion affected with tin same sign, it may be suppressed.
SIMPLE EQUATIONS. 93
1. In the equation 2x — 12 4 '3 = a; — 5a + 9, transpose
the unknown terms to the first member, and the known terms
to the second.
Eesult, 2x — x + 5 a: = 12 — 3 + 9.
EXAMPLES.
Transpose the unknown terms to the first member, and the
known terms to the second, in the following :
2. 3x — 2a = 45+.2x.
3. 4:X + 9 = 2512x.
4. 4 a 2 x + b 2 = — 4 a bx+±ac+ b 2 .
5. a c + c x — a d = 2 a — 7 #.
6. & c + a 2 x — m ?r = b x + a tZ — 5.
7. 3 — & — x = c — 3x.
8. 2a — 3c = 5ic — b — dx.
9. 10 jc  312 = 32 x + 21  52 x.
CLEARING OF FRACTIONS.
175. 1. Clear the equation jr T = X+ s of fractions.
o 4 o o
The least common multiple of 3, 4, G, and 8 is 24. Multi
plying each term of the equation by 24 (Art. 170), we have
16 x  30 = 20 x + 9,
where the denominators have been removed. Hence the fol
lowing
RULE.
Multiply each term of the equation by the least common
multiple of the denominators.
2,
ax (I x m
c —
b en
x 2 a 1 x
:;.
2 a 3 b lab 6
ax ex a
4.
x i + , — o
b a e
94 ALGEBRA.
Note. The operation of clearing of fractions may be performed by
multiplying each term of the equation by any common multiple of the de
nominators. The product of all the denominators is obviously a common
multiple, and the rule is sometimes given as follows : "Multiply each term
of the equation by the product of all the denominators."
EXAMPLES.
Clear the following equations of fractions:
3x 5x b
4 3
7. x %+ 20 = 1 + ^ + 26.
n w X O ox n
8 ' is^ir +! " l = () 
—, *As its %K/ C\C\ f\ *^ *^ *^ HO JO </•
b ' 5 + L2 = l0 _ " " 12" 3T~ lo = 8~~6~
5 x 5 \^ q x 9J i_j x
10. Clear the equation 21 — ■ — == — — —   t
o 1G 2
of fractions.
The least common denominator is 16 ; multiplying each term
by 1G, we have
336  (10 x  10) = 11  3 x  (776  56 x)
or, : !: U ;  lO x + 10 = 11  3 x  776 + 56 x, Am.
Note. When a fraction, whose numerator is not a monomial, is preceded
by a —sign, it will he found convenient, on clearing of fractions, to enclose
the numerator in a parenthesis. If this is not done, care must be taken
to change the sign of each term in the numerator.
Clear the following equations of fractions :
11 x a + x 15 12 ax + b cx + d a
2 3~~ ~~2' r be ' V
SIMPLE EQUATIONS. 95
13. 1 1^ = 0. 15. ?_ ?___5a!_
1 + a; 1 — a? x + 1 a; — 1 ar — 1
14 a ar — 3 1 _a 16 ,T ~*~ ^ a? — 3 2« + l_
2~2x + l _ 3 _ ' * ~5~ ~2~ ~3~
CHANGING SIGNS.
176. The signs of all the terms of an equation may be
changed without destroying the equality.
For, in the equation a — x = b — c, let all the terms he
multiplied by — 1 (Art. 170). Then,
— a + x = — b + c
or, x — a = c — b.
For example, the equation — 5 x — a = 3 x — b, by chang
ing the signs of all the terms, may he written
5x + a = b — 3x.
SOLUTION OF SIMPLE EQUATIONS.
177. To solve a simple equation containing hut one un
known quantity.
1. Solve the equation 5 x — 7 = a? + 9.
Transposing the unknown terms to the first member, and
the known terms to the second,
5 x — ^ = 7 + 9
Uniting similar terms, 4 x = 16
Dividing each member by 4 (Art. 170),
x = 4, Ans.
This value of x we may verify (Art. 168). Thus, substi
tuting 4 for x in the given equation, it becomes
20  7 = 4 + 9,
which is identical ; hence the value of x is verified.
96 ALGEBRA.
2. Solve the equation 8 x + 19 = 25 x — 32.
Transposing, 8 x — 25 x = — 19 — 32
Uniting terms, — 17 x = — 51
Dividing by — 17, cc = 3, Ans.
To verify the result, put 3 for x in the given equation.
Then, 24 + 19 = 75  32
or, 43 = 43.
o a i .i . 3x 5 2x x
o. bolve the equation — —   = — .
4 o *j
Clearing of fractions, by multiplying each term of the equa
tion by 12, the least common multiple of the denominators,
9x + 10 = 8x — Gx
Transposing, 9a; — 8x + 6x = — 10
Uniting terms, 7 x = — 10
Dividing by 7, x = — — , Ans.
To verify this result, put x = =r in the given equation.
Then, _30 5_ _20 10
_ .28 + 6~  21 + 14
or,
or,
90 + 70 _80 + 60
~84 _ "SlT
_20_ _20
84 " 84'
RULE.
Clear the equation of fractions if it has any. Transpose
the unknown terms to tit e first member, and the known terms
to the second, and reduce each member to its simples} firm.
Diride both members of the resulting equation by the coefficient
of the unknown quantity.
SIMPLE EQUATIONS. 97
EXAMPLES.
Solve the following equations :
4. 3sc + 5 = a; + ll. 7. 3x + 25ce = :b7 + 3.
5. 3z2:=5xlG. 8. 185cc2x = 3 + a; + 7x.
6. 22a; = 3a?. 9. 5x^3 + 17 = 192x2.
10. Solve the equation
5(7 + 3x)(2a;3)(l2x)(2a ; 3) 2 (5 + : *0=O
Performing the operations indicated, we have
35 + 15a: + 4a; 2 8:r + 34x 2 + 12:r95a; =
Transposing, and suppressing the terms 4 x 2 and — 4 x 2 ,
15 ;c8;c + 12 a:a; =  35 3 + 9 + 5
18 x =  24
24 4 ,
x = lS = 3> AnS 
Solve the following equations :
11. 3 + 2 (2x + S) = 2x 3(2 x + 1).
12. 2cc — (4a;l)=5a;0l).
13. 7 («2) 5 (a + 3) = 3 (2x 5) 6 (4a; 1).
14. 3(3x + 5)2(5z3)=13(5;c16).
15. (2 x  1) (3 a; + 2) = (3 a;  5) (2 a + 20).
16. (5  G a) (2 x  1) = (3 x + 3) (13  4 x).
17. (^3)' 2 (5a) 2 = 4'a;.
18. (2a;l) 2 3(^2) + 5(3x2)(52a ; j 2 = 0.
• 3 7 7 5
19. Solve the equation „ — = zr^ — k~ •
98 ALGEBRA.
Clearing of fractions, by multiplying each term by 12 x, the
least common multiple of the denominators,
36  42 = 7 x  20
 7 x =  36 + 42  20
 7 a; =  14
Solve the following equations :
20. i^_7=— — 24 ^— x 2x 8x 11
4 3 4 ' * 5 *^2— 11.
2i 1, 1 1 1 ok x ^ x _x 3x
6 + 2^ _ 4 + 12^ Mt 2 + ~6~3 = 6~7T
«.  +  = 18. 26.*f + 20 =  +  + 2 6.
23. _?_*=I_i 27. 2^ = 7 3
345a? a; 2a; 2a;'
oqqt n . • 3a; — 1 2 a; + 1 4a; — 5
ao. holve the equation : = 4.
4 3 5
Multiplying each term by 60,
45 x — 15  (40 x + 20)  (48 x  60) = 240
45 x — 15  40 x — 20 — 48 x + 60 = 240
45 x  40 x — 48 x = 15 + 20  60 + 240
 43 a; = 215
X = — 5, ^1?2S.
Solve the following equations :
oq q 5 ./■ + 3 7x OA 2 .r + 1 r 5
29. 3>x\ — = s. 30. x =— = 5x — =.
72 5 3
31. 7 a 7 = 3 x + 7.
SIMPLE EQUATIONS. 99
32. 2 7 ^~^ = 3x
33.
34.
6 4
5a2_3a + 4 7x\2 _x — 10
"IT ~T~ ~6~ ~2~
a; + 1 2 a; — 5 _ 11 x + 5 a; — 13
~2~ "5" ~W ~3 '
5a + l 17 x + 7 3.T1 7 a; — 1
4 + a 3 a; — 2 11 a + 2 2  9 a;
36.
14
2 a + 1 4 a + 5 8 + x 2x + 5
' ~3~~ ~T~ ~~6 _ ~8~
2 3 1
38. Solve the equation
x — 1 a 1 + 1 a'~ — 1 "
Clearing of fractions, by multiplying each term by x 2 — 1,
2 (a + 1)  3 (x  1) = 1
2a; + 23a + 3 = l
2a3a; = 23 + l
— a: = — 4
x = 4, ^4«s.
4 a + 3 12 a  5 2 x  1
39. Solve the equation
10 5 a; — 1 " 5
Clearing of fractions partially, by multiplying each term
by 10,
120 a; 50 ,
4a; + 3 = — t— = ±x — 2
o .'' — 1
_ , . 120 a; 50
4a; + 3 — 4.r + 2 = — = .—
ox — 1
: 120 x  50
5 = —= —
o a — 1
100 ALGEBRA.
Clearing of fractions, by multiplying each term by 5 x — 1,
25 x  5 = 120 x  50
25 a;  120 x = 5  50
 95 £ =  45
_45_ 9
x y5vJ> Ans '
Note. If the denominators are partly monomial, and partly polynomial,
clear of fractions at first partially, multiplying by such a quantity as will
remove the monomial denominators.
Solve the following equations :
1 —X 1 + x 1 — ar
x — 1 x + 1 3
X X 2 —
5x 2
3 j3 x 
7 3'
2xl
2* + 7
3a; + 4~
"3^ + 2
52a
32x
41 — ' 4fj
* ' x ._ 2 s + 2~x 2 4'
' ~x~+Y'' : ^+T* 9 = "3" " 1dx '
„_ <6x 2 3x + 2 _ ._ 2ar + 3a; 1
43  rr^ s s = 3. 47. ^ — — + —  = x + 1.
2 a; 2 + 5 a; — 7 2 * + 1 3 <c
48. Solve the equation 2 a ce — 3b = x + c — 3 ax.
Transposing and uniting terms, 5 a x — x = 3b \ c
Factoring the first member, x (5 a — 1) = 3 b + c
Dividing by 5 a — 1, a; = = r , ^4»s.
49. Solve the equation (& — c .r) 2 — (a — c x)' 2 = b (b — a).
Performing the operations indicated,
b 2 2bcx+ c 2 x 2  a 2 + 2acx c 2 x 2 = b 2  a b
Suppressing the term lr in both members, and the terms
c 2 x 2 and — c 2 x 2 in the first member,
SIMPLE EQUATIONS. 101
— 2bcx — a 2 + 2acx = — ab
2 a ex — 2b cx = a 2 — ab
Factoring both members, 2 c x (a — b) = a (a — b)
n . ,. a (a — b) a .
Dividing by 2e(ab), x= 2c ^_^ = ^ , Ans.
Solve, the following equations :
50. 2 ax + d — 3c — bx.
51. 2 x — Aa = 3 ax + a 2 — a 2 x.
52. 2 a x + 6 b 2 = 3 b x + 4 a b.
53. 6 b m x — 5 a n — lo a ni — 2bnx.
54. (or  2 x) 2 = (4 x  b) (x + 4 e).
55. (2 a  3 x) (2 a + 3x) = b 2 (3x b) 2 .
56. (3 a — x) (a + 2 x) = (5 a + x) (a — 2 x),
3b x 2 _ 3 _ 2bx
c a c
 3 +4^=3V 2 «( 2  3 «>
57.
a
X
58.
2a
59.
X
2
60.
X
a b
61.
X
2
1 + 2 ax 2x + \
2 a a 2
X + " b X
~~3lT = 3T~
(«!)•
I> c x x a c — 4 6 a;
2b c 6 c 3b c
62. Solve the equation .2 x  .01  .03 x = .113 x + .161.
FIRST METHOD.
('banging the decimals into common fractions.
2x 1 3x 118 a; 161
10 100 100 " 1000 1000
102 ALGEBRA.
Multiplying each term by 1000,
200 x  10  30 x = 113 x + 161
57 x = 171
x = 3, Ans.
SECOND METHOD.
Transposing, .2 x — .03 x — .113 x = .01 + .101
Uniting terms, .057 x = .171
Dividing by .057, x = 3, ^4ms.
Solve the following equations :
63. .3x .02  .003 x = .7 .06 a  .006.
64. .001 x  .32 = .09 x  .2 x  .653.
65. .3 (1.2 X 5)=U + .05 x.
66. .7 (x + .13) = .03 (4x .1) + .5.
67. 3.3a; \ =.la; + 9.9.
.5
23a 5a: _ 2x 3 _ a  2 7
"T5 — + L25~ ~9~ "378 ^ " 9 '
178. To prove that a simple equation ran have but one root.
We have soon that every simple equation can be reduced to
the form x = a.
Suppose, if possible, that a simple equation can have two
roots, and that >\ and r., are the roots of the equation x = a.
Then (Art. 168),
r x — a,
r 2 = a.
Hence, t x = r 2 \ that is, the two supposed roots are identical.
Therefore a simple equation can have but one root.
PROBLEMS. 103
XIII. — PROBLEMS
LEADING TO SIMPLE EQUATIONS CONTAINING ONE
UNKNOWN QUANTITY.
179. A Problem is a question proposed for solution.
180. The Solution of a problem by Algebra consists of
two distinct parts :
1. The Statement, or the process of expressing the condi
tions of the problem in algebraic language, by one or more
equations.
2. The Solution of the resulting equation or equations, or
the process of determining from them the values of the un
known quantities.
The statement of a problem often includes a consideration
of ratio and proportion (Art. 21).
181. Ratio is the relation, with respect to magnitude,
which one quantity bears to another of the same kind, and is
the result arising from the division of one quantity by the
other.
A Proportion is an equality of ratios.
Thus,
a : b, or  , indicates the ratio of a to b.
a : b = c : d, is a proportion, indicating that the ratio of a
to b, is equal to the ratio of c to d.
In a proportion the relation of the terms is such that the
product of the first and fourth is equal to the product of the
second and third.
ct c
For, a : b = c : <% is the same as j — , which, by clearing of
fractions, gives ad = b c.
1 04 ALGEBRA.
182. For tlie statement of a problem no general rule can
be given ; much must depend on the skill and ingenuity of the
operator. We will give a few suggestions, however, which
will be found useful :
1. Express the unknoivn quantity, <>r one of the unknoivn
quantities, by taw of the final letter* of the alphabet.
1'. From tlie given conditions, find expressions for the other
unknown quantities, if any, in the problem.
3. Form on equation, by indicating the operations necessary
to verify the values of the unknown quantities, were they
already known.
4. Determine the value of the unknown quantity in the
equation th US formed.
Note. Problems which involve several unknown quantities may often
be solved by representing one of them only by a single unknown letter.
1. What number is that to which if four sevenths of itself
be added, the sum w.ill equal twice the number, diminished by
27?
Let x = the number.
4 x
Then =— = four sevenths of it,
and 2x = twice it.
4 x
By the conditions, x \ — — = 2 x — 27
Solving this equation, x = 63, the number required.
2. Divide 144 into two parts whose difference is 30.
Let x = one part.
Tli en. 144 — x = the other part.
By the conditions, x — (144 — x) = 30
Solving this equation, x = 87, one part.
144 — #= ~>7, the other part.
PROBLEMS. 105
3. A is three times as old as B ; and eight years ago he was
seven times as old as B. What are their ages at present ?
Let x = B's age.
Then, 3 x — A's age.
Now, x — 8 = B's age, eight years ago,
and 3 x — 8 = A's age, eight years ago.
By the conditions, 3x — 8 = 7 (x — 8)
Whence, x = 12, B's age,
and, 3 x = 36, A's age.
4. A can do a piece of work in 8 days, which B can perform
in 10 days. In how many days can it he done hy both work
ing together ?
Let x = the number of days required.
Then,  = what both can do in one day.
Also, — = what A can do in one day,
o
and j = what B can do in one day.
Since the sum of what each separately can do in one day is
equal to what both can do together in one day,
i JL i
8 + 10 _ x
Whence, x = 4f , number of days required.
5. A man has $ 3.64 in dimes, halfdimes, and cents. He
has 7 times as many cents as half : dimes, and one fourth as
many halfdimes as dimes. How many has he of each ?
106 ALGEBRA.
Let x = the number of dimes.
x
Then,  = the number of halfdimes,
4
7 x
and — r— = the number of cents.
4
Now, 10 x = the value of the dimes in cents,
and —j— == the value of the halfdimes in cents.
4
By the conditions, 10 x \ ; —  — = 364
J '44
Whence, x = 28, number of dimes,
x
 = 7, number of halfdimes,
4
— t— = 49, number of cents.
6. Two pieces of cloth were purchased at the same price per
yard ; but as they were of different lengths, the one cost $ 5
and the other $ 6.50. If each had been 10 yards longer, their
lengths would have been as 5 to 6. Required the length of
each piece.
Since the price of each per yard is the same, the lengths of
the two pieces must be in the ratio of their prices, that is, as 5
to 6h, or as 10 to 13. Therefore,
Let 10 x = the length of the first piece in yards,
and 13 x = the length of the second piece in yards.
By the conditions, 10 x + 10 : 13 x + 10 = 5 : 6
or (Art. 181), 6 (10 x + 10) = 5 (13 x + 10)
"Whence, x = 2.
Then, 10 x = 20, length of first piece,
and 13 x = 26, length of second piece.
PROBLEMS. 107
7. The second digit of a number exceeds the first by 2 ; and
if the number, increased by 6, be divided by the sum of the
digits, the quotient is 5. Required the number.
Let x = the first digit.
Then, x + 2 = the second.
Since the number is equal to 10 times the first digit, plus
the second,
10 x + x + 2, or 11 x + 2 = the number.
11 x __ 2 + 6
By the conditions, ■ ^— = 5
J x + x + 2
Whence, x = 2, the first digit,
and x + 2 = 4, the second digit.
Therefore the number is 24.
8. Two persons, A and B, 63 miles apart, set out at the
same time and travel towards each other. A travels 4 miles
an hour, and B 3 miles. What distance will each have trav
elled when they meet ?
Let x = the distance A travels.
Then, 63 — x = the distance B travels.
x
 = the time A takes to travel x miles,
and — — — = the time B takes to travel 63 — x miles.
o
By the conditions of the problem, these times are equal ;
x 63 — x
4=^r
Whence, x = 36, A's distance,
and 63 — x = 27, B's distance.
108 ALGEBRA.
9. At what time between 3 and 4 o'clock are the hands of a
watch opposite to each other ?
Let M represent the position of
the minutehand at 3 o'clock, and H
the position, of the hourhand at the
same time.
\jj Let M 1 represent the position of
Ih' the minutehand when it is opposite
to the hourhand, and H 1 the po
sition of the hourhand at the same
time.
Let x = the arc M H H' M', the space over which the min
utehand has moved since 3 o'clock.
x
Then, ^ = the arc H H', the space over which the hour
hand has moved since 3 o'clock.
Also, the arc MH= 15 minute spaces,
and the arc H' M 1 = 30 minute spaces.
Now, arc M H H 1 M = arc MH+ arc H H< + arc H< M,
x
or, x = 15 + j^ + 30
Solving this equation, x = 49 ^ minute spaces.
That is, the time is 49^ minutes after 3 o'clock.
PROBLEMS.
10. My horse and chaise are worth $ 336 ; but the horse is
worth twice as much as the chaise. Required the value of
each.
11. What number is that from which if 7 be subtracted, one
sixth of the remainder will be 5?
12. What two numbers are those whose difference is 3, and
the difference of whose squares is 51 ?
PROBLEMS. 109
13. Divide 20 into two such parts that 3 times one part may
be equal to one third of the other.
14. Divide 100 into two parts whose difference is 17.
15. A is twice as old as B, and 10 years ago he was 3 times
as old. What are their ages ?
16. A is four times as old as B ; in thirty years he will be
only twice as old as B. What are their ages ?
17. A can do a piece of work in 3 days, and B can do the
same in 5 days. In how many days can it he done by both
working together ?
18. A can do a piece of work in 3§ hours, which B can do
in 2 hours, and C in 2i hours. In how many hours can it be
done by all working together ?
19. A and B can do a piece of work together in 7 days,
which A alone can do in 10 days. In what time could B alone
do it ?
20. The first digit of a certain number exceeds the second
by 4; and when the number is divided by the sum of the
digits, the quotient is 7. What is the number '.'
21. The second digit of a certain number exceeds the first
by 3; and if the number, diminished by 9, be divided by the
difference of the digits, the quotient is 9. What is the
number ?
22. A drover has a lot of oxen and cows, for which he gave
$ 1428. For the oxen he gave $ 55 each, and for the cows $ 32
each ; and he had twice as many cows as oxen. Required the
number of each.
23. A gentleman, at his decease, left an estate of $1872 for
his wife, three sons, and two daughters. His wife was to re
ceive three times as much as either of her daughters, and each
son to receive one half as much as each of the daughters. Re
quired the sum that each received.
HO ALGEBRA.
24. A laborer agreed to serve for 36 days on these condi
tions, that for every day he worked he was to receive $1.25,
but for every day lie was absent he was to forfeit * 0.50. At
the end of the time he received $ 17. It is required to find
, how many days he labored, and how many days he was absent.
25. A man, being asked the value of his horse and saddle,
replied that his horse was worth $114 more than his saddle,
and that g the value of the horse was 7 times the value of the
saddle. What was the value of each ?
26. In a garrison of 2744 men, there are 2 cavalry soldiers
to 25 infantry, and half as many artillery as cavalry. Re
quired the number of each.
27. The stones which pave a square court would just cover
a rectangular area, whose length is 6 yards longer, and breadth
4 yards shorter, than the side of the square. Find the area of
the court.
28. A person has travelled altogether 3036 miles, of which
he has gone 7 miles by water to 4 on foot, and 5 by water to
2 on horseback. How many miles did he travel in each
manner ?
29. A certain man added to his estate ^ its value, and then
lost $ 760 ; but afterwards, having gained $ 600, his property
then amounted to $ 2000. What was the value of his estate at
first?
30. A capitalist invested § of a certain sum of money in
government bonds paying 5 per cent interest, and the re
mainder in bonds paying 6 per cent ; and found the interest
of the whole per annum to be $180. Required the amount of
each kind of bonds.
31. A woman sells half an egg more than halt her eggs.
Again she sells half an egg more than half her remaining
eggs. A third time she does the same; and now she has sold
all her eggs. How many had she at first ?
PROBLEMS. HI
32. What number is that, the treble of which, increased by
12, shall as much exceed 54, as that treble is less than 144 ?
33. A ashed B how much money he had. He replied, " If
I had 5 times the sum I now possess, I could lend you $ 60,
and then i of the remainder would be equal to h the dollars I
now have." Required the sum B had.
34. A, B, and C found a purse of money, and it was mutu
ally agreed that A should receive $ 15 less than one half, that
B should have $13 more than one quarter; and that C should
have the remainder, which was $ 27. How many dollars did
the purse contain?
35. A number consists of 6 digits, of which the last to the
left hand is 1. If tins number is altered by removing the 1
and putting it in the units' place, the new number is three
times as great as the original one. Find the number.
36. A prize of $ 1000 is to be divided between A and B, so
that their shares may be in the ratio of 7 to 8. Required the
share of each.
37. A man has $ 4.04 in dollars, dimes, and cents. He has
one fifth as many cents as dimes, and twice as many cents as
dollars. How many has he of each ?
38. I bought a picture at a certain price, and paid the same
price for a frame ; if the frame had cost $ 1.00 less, and the
picture $ 0.75 more, the price of the frame would have been
only half that of the picture. Required the cost of the
picture.
39. A gentleman gave in charity $ 46 ; a part in equal por
tions to 5 men, and the rest in equal portions to 7 women.
Now, a man and a woman had between them $8. What
was given to the men, and what to the women ?
40. Separate 41 into two such parts, that one divided by
the other may give 1 as a quotient and 5 as a remainder.
112 ALGEBRA.
41. A vessel can be emptied by three taps ; by the first
alone it could be emptied in 80 minutes, by the second in 200
minutes, and by the third in 5 hours. In what time will it be
emptied if all the taps be opened ?
42. A general arranging his troops in the form of a solid
square, finds he has 21 men over; but, attempting to add
1 man to each side of the square, finds he wants 200 men to
fill up the square. Required the number of men on a side at
first, and the whole number of troops.
43. At what time between 7 and 8 are the hands of a watch
opposite to each other ?
44. At what time between 2 and 3 are the hands of a watch
opposite to each other?
45. At what time between 5 and 6 are the hands of a watch
together ?
46. Divide 43 into two such parts that one of them shall be
3 times as much above 20 as the other wants of 17.
47. Gold is 19} times as heavy as water, and silver 10i
times. A mixed mass weighs 4160 ounces, and displaces 250
ounces of water. What proportions of gold and silver does it
contain ?
48. A gentleman let a certain sum of money for 3 years at
5 per cent compound interest ; that is, at the end of each year
there was added J,, to the sum due. At the end of the third
year there was due him .$2315.25. Required the sum let.
49. A merchant ha!s grain worth 9 shillings per bushel 3 and
other grain worth 1.'! shillings per bushel, in what proportion
must he mix 40 bushels, so that he may sell the mixture at
10 shillings per bushel '.'
50. A alone could perform a piece of work in L2 hours; A
and C together could do it in 5 hours; and C's work is § of
B's. Now. the work has to be completed by noon. A begins
work at 5 o'clock in the morning; at what hour can he he
relieved by B and ( '. ami the work he just finished in time'.'
SIMPLE EQUATIONS. H3
51. A merchant possesses $5120, but at the beginning of
each year he sets aside a fixed sum for family expenses. His
business increases his capital employed therein annually at the
rate of 25 per cent. At the end of four years he finds that his
capita] is reduced to $3275. What are his annual expenses?
52. At what times between 7 and 8 o'clock are the hands of
a watch at right angles to each other ?
53. At what time between 4 and 5 o'clock is the minute
hand of a watch exactly five minutes in advance of the hour
hand ?
54. A person has 11^ hours at his disposal ; how far may
he ride in a coach which travels 5 miles an hour, so as to re
turn home in time, walking back at the rate of oh miles an
hour ?
55. A fox is pursued by a greyhound, and is 60 of her own
leaps before him. The fox makes 9 leaps while the greyhound
makes but 6 ; but the latter in 3 leaps goes as far as the former
in 7. How many leaps does each make before the greyhound
catches the fox ?
56. A clock has an hourhand, a minutehand, and a second
hand, all turning on the same centre. At 12 o'clock all the
hands are together, and point at 12. How long will it be
before the minutehand will be between the other two hands,
and equally distant from each ?
XIV. — SIMPLE EQUATIONS
CONTAINING TWO UNKNOWN QUANTITIES.
183. If we have a simple equation containing two unknown
quantities, as 3 x — 4 y = 2, we cannot determine definitely
the values of x and y ; because, for every value which we give
to one of the unknown quantities, we can find a corresponding
114 ALGEBRA.
value for the other, and thus find any number of pairs of values
which will satisfy the given equation.
Thus, if we put x = G, then 18 — 4 y = 2, or y = 4 ;
t if we put x = — 2, then — 6 — 4 y = 2, or y = — 2 ;
if we put a; = 1, then 3 — 4 ?/ = 2, or y = £ ; etc.
And any of the pairs of values < " , I, ■! n i, ■< ~ , ,
etc., will satisfy the given equation.
If we have another equation of the same kind, as 5x + 7?/=17,
we can find any number of pairs of values which will satisfy
this equation also.
Now suppose we are required to determine a pair of values
which will satisfy both equations. We shall find but one pair
of values in this case. For, multiply the first equation by 5 ;
thus,
15 x 20 y = 10;
and multiply the second equation by 3 ; thus,
15 x + 21 y = 51.
Subtracting the first of these equations from the second
(Art. 44), we have
41 y = 41,
or, p = l.
In the first given equation put y = l; then 3 x — 4 = 2, or
3 x = 6 ; whence, x = 2. The pair of values \ ~i\ satisfies
both the given equations; and no other pair of values can be
found which will satisfy both.
184. Simultaneous Equations are such as are satisfied by
the same values of their unknown quantities.
185. Independent Equations are such as cannot be made
to assume the same form.
SIMPLE EQUATION'S. 115
186. It is evident, from Art. 183, that two unknown
quantities require for their determination two independent,
simultaneous equations. When two such equations are given,
it is our object to obtain from them a single equation contain
ing but one unknown quantity. The value of that unknown
quantity may then be found; and by substituting it in either
of the given equations we can find, as in Art. 183, the value of
the other.
ELIMINATION".
187. Elimination is the process of combining simultaneous
equations so as to obtain from them a single equation contain
ing but one unknown quantity.
There are four principal methods of elimination : by Addi
tion or Subtraction, by Substitution, by Comparison, and by
Undetermined Multipliers.
CASE I.
188. Ellin {nation by Addition or Subtraction.
1. Given ox — 3 y = 19, and 7 x + 4 y — 2, to find the
values of x and y.
Multiplying the first equation by 4, 20 x — 12 y = 76
Multiplying the second equation by 3, 21 x + 12 y = 6
Adding these equations, 41 x = 82
Whence, x = 2.
Substituting this value in the first given equation,
103y = 19
3y = 9
y = 3.
We might have solved the equations as follows :
Multiplying the first by 7, 35 x  21 y = 133 (1)
Multiplying the second by 5, 35 x + 20 y = 10 (2)
Subtracting (2) from (1), — 41 y = 123
2/ = 3.
116 ALGEBRA.
Substituting this value of y in the first given equation,
5 x + 9 = 19
5 a = 10
x = 2.
The first of these methods is elimination by addition ; the
second, elimination by subtraction.
RULE.
Multiply the given equations, if necessary, by such numbers
or quantities as will make the coefficient of one of the unknown
quantities the same in the two resulting equations. Then, if
the signs of the terms having the same coefficient arc alike,
subtract one equation from the other ■ if unlike, add the two
equations.
This method of elimination is usually the best in practice.
CASE II.
189. Elimination by Substitution.
Taking the same equations as before,
5 x — 3 y = 19 (1)
7x + 4t/= 2 (2)
Transposing the term 7 x in (2), 4 y = 2 — 7 x
2 7 x
Dividing by 4, y = — _ (3)
Substituting this value of y in (1),
5*_3 (^=^) =19
Performing the operations indicated,
SIMPLE EQUATIONS. 117
Clearing of fractions, 20 x — (6 — 21 x) = 76
or, 20 x  6 + 21 x = 76
Transposing, and uniting terms, 41 x = 82
Whence, x = 2.
2 — 14
Substituting this value in (3), y = — j — = — 3.
• RULE.
7v//r/ £/ie 7v//»e o/o/^e o/ £Ae unknown quantities in terms
of the other, from cither of the given equations; and substi
tute this value for that quantity in the other equation.
This method is advantageous when either of the unknown
quantities has 1 for its coefficient.
CASE III.
190. Elimination by Comparison.
Taking the same equations as before,
5 x  3 y = 19 (1)
7x + ±y= 2 (2)
Transposing the term — 3 y in (1), 5 x = 3 y + 19
3//+ 19 ,,.
or, a = — g (3)
Transposing the term 4 y in (2), 7 a; = 2 — 4 y
or, a: = —
Placing these two values of x equal to each other (Art. 44),
3// + 19 _ 24y
5 7
Clearing of fractions, 21 y + 133 = 10 — 20 y
118 ALGEBRA.
Transposing, and uniting terms, 41 y = — 123
Whence, y — — 3.
— 9 + 19
Substituting this value in (3), x = p
o
RULE.
Find the value of the same unknown quantity in terms of
the other, from each of the given equations : ami form a new
equation by placing these values equal to each other.
CASE IV.
191. Elimination by Undetermined Multipliers.
An Undetermined Multiplier is a factor, at first undeter
mined, but to which a convenient value is assigned in the
course of the operation.
Taking the same equations as before,
5x3y = 19 (1)
7 x + 4 y = 2 (2)
Multiplying (1) by m, 5 m x — 3 m y = 19 m (3)
Subtracting (3) from (2),
7 x — 5mx + iy + 3m y — 2 — 19 m
Factoring, x (7 — 5 m) + y (4 + 3 m) =2 — 19 m (4)
Now, let the coefficient of y, 4 + 3 m = ; then 3 m = — 4,
4
or m = — Kj substituting this value of m in (4),
o
/_ 20\ „ 76
n 7 +ir) = 2 +3
Clearing of fractions, x (21 + 20) = 6 + 70
41 x = 82
x = 2.
SIMPLE EQUATIONS. 119
Substituting this value in (2), 14 + 4 y = 2
4y = 12
y = 3.
We might liave let the coefficient of x in (4), 7 — 5m = 0;
7
then m would have been ■= ; substituting this value of m in (4),
o
y(±+ )=2
Clearing of fractions, y (20 + 21) = 10  133
41 y =  123
y = 3.
Instead of subtracting (3) from (2), we migbt have added
them and obtained the same results. Also, in the first place,
we might have multiplied (2) by m, and either added the re
sult to, or subtracted it from, (1).
RULE.
Multiply one of the given equations by the undetermined
quantity, m ; and add the result to, or subtract it from, the
other given equation.
In the resulting equation, factored with reference to the
unknown quantifies, place the coefficient of one of the un
known quantities equal to zero, and find the value of ra.
Substitute tins value of m In the equation, and the result trill
be a simple equation containing but one unknown quantity.
This method is advantageous in the solution of literal
equations.
2. Solve the equations.
ax + b y — c (1)
a f x + b'y = c' (2)
120 ALGEBRA.
Multiplying (1) by m, a m x + b m y = c m (3)
Add (2) and (3), a' x + a m x + V y + b m y = d + cm
Factoring, x (a' + a vi) + y (b 1 + b m) = d + c m (4)
In (4), put the coefficient of y, V + b in, equal to zero.
Then, b m — — V ; whence, m = — .
b
Substituting this value of m in (4),
a V \ , c b
x \a' — ) = d
Clearing of fractions, x (a' b — ab') =b d — b' e
1, ( j _ y c
Whence, x = — — .
a' b — ab'
In (4), put the coefficient of x, a' + a m, equal to zero.
Then, a m = — a ' ; whence, m = ■ .
a
Substituting this value of m in (4),
a'b\ a'e
y\ h — ') — c —
a ' a
Clearing of fractions, y (ab 1 — a 1 b) = ad — a' c
a d — a' c
Wh ence, y = —r l — .
ab' — a' b
Before applying either of the preceding methods of elimina
tion, the given equations should be reduced to their simplest
forms.
EXAMPLES.
192. Solve, by whichever method may be most advanta
geous, the following equations :
3. 3cc + 72/ = 33; 2x + ±y = 20.
4. 7x + 2y = 31; 3 z ~ 4 ?/ = 23.
5. 6x3y = 27; 4;r6y = 2.
SIMPLE EQUATIONS. 121
6. 7 x + 3y = 50; 2y5x = U.
7. 8y + 12 .£ = 116 ; 2x — y = 3.
8. 11 x + 3 y =  124 ; 2 x  6 y = 56.
9. 9a; + 4?/ = 22;27/ + 3cc = 14.
10. ^+^f8; 8x + 2y = S0.
11. 7z2y = G;2a; + 22/ = 24.
,«. ^ ,.„ 2^; 11?/ 5
12. 11 y + G a; = 115 ; — — = — ■= .
13.  a} + y = ^ ; 10*12y = 62.
5 7
14. _7a; + 47/ = 113; £ + ?, = .
15> 2~3°' 4 + G b 
16. ^ y = 31; a; + ^ = 33.
17. A^_^ = _30; aj + 7y = 119.
< 3 a
18. rc + 2// = .G; 1.7 x y = . 58.
3 ^ ?/ cc 2 ?/
19. t r ?? 1[g L=^.4y8 aJ = lL
2 T
» + 3y 3 7 y  x
' 2xy 8' 2 + x + 2y ~
21. a x + b y = m ; c x + d y= n.
22. m cc + ?i ?/ = r ; m' cc — n' y = /.
no xy y x
aw a c
122 ALGEBRA.
x v 1 x y 1
** a + b^ a~b a 2 b 2> ab^ a + b a 2 b 2
9* X 19 Vm*. X S 2 V~ X Q7 X + V
25l 2~ 12 = 4 +8 ' 3~ 8 4T~~^ V
2 2,3 2
26. + = 1;—  = 0.
03 + // X — V/ 03 + 2/ ^ — i7
2x + y _17 2y + 03 4 2 a?  y _ 2 y  a?
4*. x g  12 4 "' 3 4 "y " 3 '
2 03 3.7/ x + 2?/_ 5x — 6y
28< "3 5 4~~ ~° 4~"'
03 ox — y H x
29. Solve the equations,
6_3_
x y~
8 15
 + — = 1
« y
Multiplying the first equation hy 5,
30_15 = 2Q
x y '
Adding this to the second given equation,
" = 19
03
Clearing of fractions, 38 = 19 03
Whence 03 = 2.
Substituting this value in the first given equation,
y
Transposing, = 1
Whence, y = — 3.
SIMPLE EQUATIONS. 123
Solve the following equations :
3 15 2 3
1.
30.
x y 4 ' x y
31.
12 18 42 15
8
17
x y 5 ' x
y
3
32.
11 7 3 4 8
x y 2 ' x y
10.
nn a b c d
66.  +  = m ; + = ».
x y x y
fi 4 9 8
34. — + f = 4a5; ^ — = 3a 2 4J a .
<za; 6y 6a; ay
oc m ?i n m ,o
oO. 1 =zm + n; —  = iiv + n".
nx my x y
XV. — SIMPLE EQUATIONS
CONTAINING MORE THAN TWO UNKNOWN QUANTITIES.
193. If we have given three independent, simultaneous
equations, containing three unknown quantities, we may coin
hine two of them hy the methods of elimination explained in
the last chapter, so as to obtain an equation containing only
two unknown quantities ; we may combine the third equation
with either of the two former in the same way, so as to obtain
another equation containing the same two unknown quantities.
Then from these two equations containing two unknown quan
tities we may derive, as in the last chapter, the values of those
unknown quantities. These values being substituted in either
of the given equations, the value of the third unknown quan
tity may be determined from the resulting equation.
The method of elimination by addition or subtraction is
usually the most convenient.
124 ALGEBRA.
194. 1. Solve the equations,
8xdy7z = 36
12 x— y — 3z — 36
6x2y z = 10
Multiplying the first by 3, 21 x — 27 y  21 z = — 108 (1)
Multiplying the second by 2, 21 x — 2 y — 6 z = 72 (2)
Multiplying the third by 4, 21 x — Sy — 4tz = 10 (3)
Subtracting (1) from (2),
or,
Subtracting (3) from (2),
or,
Multiplying (5) by 3,
Adding (1) and (6),
Substituting this value in (5),
Substituting the values of y and z in the third given equation,
sc = 4.
In the same manner, if we have given n feidependent,
simultaneous equations, containing n unknown quantities, we
may combine them so as to form a — 1 equations, containing
n — 1 unknown quantities. These, again, may be combinnl
so as to form n — 2 equations, containing n — 2 unknown
quantities; and so on : the operation being continued until we
finally obtain one equation containing one unknown quantity.
RULE.
Multiply the given equations, if necessary, by such numbers
or quantities as trill make the coeffirirnt of one of the un
known quantities the same in the resulting equations. Cont
inue these equations by addition <>r subtraction, so as to form
2oy + loz = ISO
5y+ 3z= 36
(4)
6y 2z= 32
3y z= 16
(5)
9y 3z= 18
(6)
11//= 81
y = 6.
z = 2.
SIMPLE EQUATIONS. 125
a new set of equations, one less in number than before, and
containing one less unknown quantity. Continue the opera
tion with these new equations ; and so on, until an equation is
obtained containing nut unknown quantity.
Find the value of this unknown quantity. By substituting
it in either of the equations containing only two unknown
quantities, find the value of a second unknown quantity. By
substituting these values in either of the equations containing
three unknown quantities, find the value of a third unknown
quantity j and so on, until the values of all arc found.
Note. This rule corresponds only with the method of elimination by
addition or subtraction ; which, however, as we have observed before, is
the best in practice.
2. Solve the equations,
u + x + y = 6
u + x \ z = 9
u + y + z = 8
x+y+z=7
The solution may here be abridged by the artifice of assum
ing the sum of the four unknown .quantities to equal an auxil
iary quantity, s. Thus,
Let u + x + y + z = s.
Then we may write the four given equations as follows :
sz = 6 (1)
sy=9 (2)
sx = S (3)
su=7 (4)
Adding, 4 s — s = 30
Whence, s = 10.
Substituting the value of s in (1),,(~), (3), and (4), we obtain
z ■= 4, y = 1, x = 2, and u = 3.
126 ALGEBRA.
EXAMPLES.
Solve the following equations :
3. x + y+z = 53; x + 2 y + 3s = 107; x + 3y + 4s = 137.
4. 3xy — 2z = 23\ G x + 2 y + 3 z = 15;
4x + 3 y — z = — G.
5. 5x — 3y+2z = 41i 2x + yz = l7; 5x + ±y2z = 3%.
6. 7a; + 47/2 = 50; 4x 5 y3 z = 20;
x — 3y — 4:Z = 30.
7. 3u + x + 2y — z = 22; 4x — y + 3 s = 35;
4:u + 3x2y=19; 2 u + 4y + 2 2 = 46.
8. a; + ?/ = 2j a2 + s = 3; y + * = — 1.
9. ?/ + s = a ; aj + £ = J ; a; + y = c .
10. 4a;4y = a + 4s; 6y — 2a; = a + 2s; 7sy = fl + ^.
11 2 + 3~4~ 43 ' 3"I + 2° 4 ' 4 + 2~3 == ~°°
12. 2a + 2 2 , + 3 = 172w; ?/ + 3*=2; 4aj. + * = 13j
^ + 3y = 14.
13. a y \ b x = c; c x + a z = b ; b z + c y = a.
14 4 ?_ 6 81 _2_ _3_ 10 _7
a: y 2~ = 2~' 3^" + 2l/~7 r i" Z "2
8 6 4 ..
+ — = 11.
9 .'■ // 7
. 3 2 J_ J_ 1 A A 1
°' 4x "37/" ; ' ""32/ + 2z~ ; 2z + 4:x~
16. x — ay+ cr z — a 8 ; x — by+b' 2 z = b :i ; x — c y + c' 2 z = c
8
PROBLEMS. 127
17 y~ z x + % _ 1 x —y x — z —n.
2 ~ 4 ~2' 5 6
y+z x + y
4 2
4.
18. £^±1_ (2 _,) = 0; ^±^ = 2acx
a c
(« + f) 2 ac(2 + a; + s) = y.
XVI. — PROBLEMS
LEADING TO SIMPLE EQUATIONS CONTAINING MORE
THAN ONE UNKNOWN QUANTITY.
195. In the solution of problems in which we represent
more than one of the unknown quantities by letters, we must
obtain, from the conditions of the problem, as many indepen
dent equations as there are unknown quantities.
1. If 3 be added to both numerator and denominator of a
certain fraction, its value is f ; and if 2 be subtracted from
both numerator and denominator, its value is h Required
the fraction.
Let
and
By the conditions,
X
= the
numerator,
y
= the denominator,
X
+ 3
o
V
+ 3"
o
O
X
o
w
1
y
2
2
x ■■
= ~>
y = 12.
Solving these equations,
That is, the fraction is T v.
2. The sum of the digits of a number of three figures is 13 ;
if the number, decreased by 8, be divided by the sum of the
second and third digits, the result, is 25; and if 99 be added
to the number, the digits will be inverted. Find the number.
128 ALGEBRA.
Let x = the first digit,
y = the second,
and z — the third.
Then, 100 x + 10 y + z = the nuniher,
and 100 z + 10 y + x = the number with its digits inverted.
By the conditions, x + y + z = 13
100 x + 10 y + z8
— Zo
100 x + 10 y + z + 99 = 100 z + 10 y + x
Solving these equations, x = 2, the first digit,
y = 8, the second,
3, the third.
That is, the number is 283.
3. A crew can row 20 miles in 2 hours down stream, and
12 miles in 3 hours against the stream. Required the rate
per hour of the current, and the rate per hour of the crew in
still water.
Let x = rate per hour of the crew in still water,
and y = rate jjer hour of the current.
Then, x + y= rate per hour rowing down stream,
and x — y = rate per hour rowing up stream.
Since the distance divided by the rate gives the time, we
have by the conditions,
20 2
x + y
12
= 3
x—y
Solving these equations, x = 7, and y = 3.
PROBLEMS. 129
PROBLEMS.
4. A says to B, " If i of my age were added to § of yours,
the sum would be 19^ years." " But," says B, " if § of. mine
were subtracted from J of youxs, the remainder would be 18£
years." Required their ages.
5. If 1 be added to the numerator of a certain fraction, its
value is 1 ; but if 1 be added to its denominator, its value is £.
What is the fraction?
6. A farmer has 89 oxen and cows ; but, having sold 4 oxen
and 20 cows, found he then had 7 more oxen than cows. Re
quired the number of eacb at first.
7. A says to B, " If 7 times my property were added to \ of
yours, the sum would be % 990." B replied, " If 7 times my
property were added to \ of yours, the sum would be $ 510."
Required the property of each.
8. If \ of A's age were subtracted from B's age, and 5 years
added to the remainder, the sum would be 6 years ; and if 4
years were added to \ of B's age, it would be equal to fe of A's
age. Required their ages.
9. Divide 50 into two such parts that % of the larger shall
be equal to § of the smaller.
10. A gentleman, at the time of his marriage, found that his
wife's age was to his as 3 to 4 ; but, after they had been mar
ried 12 years, her age was to his as 5 to 6. Required their
ages at the time of their marriage.
11. A farmer hired a laborer for 10 days, and agreed to pay
him $ 12 for every day he labored, and he was to forfeit % 8
for every day he was absent. He received at the end of his
time % 40. How many days did he labor, and how many days
was he absent ?
12. A gentleman bought a horse and chaise for % 208, and i
of the cost of the chaise was equal to § the price of the horse.
What was the price of each ?
130 ALGEBRA.
13. A and B engaged in trade, A with $ 240, and B with
$96. A lost twice as much as B; and, upon settling their
accounts, it appeared that A had three times as much remain
ing as B. How much did each lose ?
14. Two men, A and B, agreed to dig a well in 10 days;
hut, having labored together 4 days, 1! agreed to finish the
job, which he did in 16 days. How long would it have taken
A to dig the whole well ?
15. A merchant has two kinds of grain, one at 60 cents per
bushel, and the other at 90 cents per bushel, of which he
wishes to make a mixture of 40 bushels that may be worth
80 cents per bushel. How many bushels of each kind must
he use '.'
16. A farmer has a box filled with wheat and rye; seven
times the bushels of wheat are 3 bushels more than four times
the bushels of rye ; and the quantity of wheat is to the quan
tity of rye as 3 to 5. Required the number of bushels of each.
17. My income and assessed taxes together amount to $ 50.
But if the income tax be increased 50 per cent, and the as
sessed tax diminished 25 per cent, the taxes will together
amount to $ 52.50. Required the amount of each tax.
18. A and B entered into partnership, and gained 8200.
Now 6 times A's accumulated stock (capital and profit) was
equal to 5 times B's original stock; and 6 times B's profit
exceeded A's original stock by $200. Required the original
stock of each.
19. A boy at a fair spent his money for oranges. If he had
got five more for his money, they would have averaged a half
cent less ; and if three less, a halfcent each more. J low many
cents did he spend, and how many oranges did he get?
20. A merchant has three kinds of sugar. He can sell 3
lbs. of the first quality, 4 lbs. of the second, and 2 lbs. of the
third, for 60 cents; or, he can sell 4 lbs. of the first quality,
1 lb. of the second, and 5 lbs. of the third, for 59 cents ; or, he
PROBLEMS. 131
can sell 1 lb. of the first quality, 10 lbs. of the second, and 3
lbs. of the third, for 90 cents. Required the price per lb. of
each quality.
21. A gentleman's two horses, with their harness, cost him
$120. The value of the poorer horse, with the harness, was
double that of the better horse; and the value of the better
horse, with the harness, was triple that of the poorer horse.
What was the value of each ?
22. Find three numbers, so that the first with half the other
two, the second with one third the other two, and the third
with one fourth the other two, shall each be equal to 34.
23. Find a number of three places, of which the digits have
equal differences in their order ; and, if the number be divided
by half the sum of the digits, the quotient will be 41 ; and, if
396 be added to the number, the digits will be inverted.
24. There are four men, A, B, C, and D, the value of whose
estates is $14,000; twice A's, three times B's, half of C's, and
one fifth of J)'s, is $16,000; As, twice B's, twice C's, and two
fifths of D's, is $18,000; and half of A's, with one third of
B's, one fourth of C's, and one fifth of D's, is $ 4000. Re
quired the property of each.
25. A and B are driving their turkeys to market. A says
to B, " Give me 5 of your turkeys, and I shall have as many
as you." B replies, " Give me 15 of yours, and then yours
will be f of mine." How man}' had each ?
26. A says to B and C, " Give me half of your money and I
shall have $ 55." B replies, " If you two will give me one
third of yours, I shall have $ 50." But C says to A and B, " If
I had one fifth of your money I should have $50." Required
the sum that each possessed.
27. A gentleman left a sum of money to be divided among
his four sons, so that the share of the eldest was \ of the sum
of the shares of the other three, the share of the second J of
the sum of the other three, and the share of the third I of the
132 ALGEBRA.
sum of the other three ; and it was found that the share of the
eldest exceeded that of the youngest by $ 14. "What was the
whole sum, and what was the share of each person?
28. If I were to enlarge my field by making it .") rods longer
and 1 rods wider, its area would be increased by 240 square
rods; but if I were to make its length 4 rods less, and its
width 5 rods less, its area would be diminished by 210 square
rods. Required the present length, width, and area.
29. A boatman can row down stream, a distance of 20 miles,
and back again in 10 hours; and he finds that he can row 2
miles against the current in the same time that he rows 3 miles
with it. Required the time in going and in returning.
30. A and B can perform a piece of work in days. A and
C in 8 days, and B and C in 12 days. In how many "lays can
each of them alone perform it ?
31. A person possesses a capital of $30,000, on which he
gains a certain rate of interest; but he owes $20,000, for
which he pays interest at another rate. The interest which
he receives is greater than that which he pays by $800. A
second person has $35,000, on which he gains the second rate
of interest ; but he owes $ 24,000, for which he pays the first
rate of interest. The sum which he receives is greater than
that which he pays by $ 310. What are the two rates of in
terest ?
32. A man rows down a stream, which runs at the rate of
o\ miles per hour, for a certain distance in 1 hour and 40 min
utes. In returning it takes him 6 hours and .'50 minutes to
arrive at a point 2 miles short of his startingplace. Find the
distance lie pulled down the stream, and the rate of his pulling.
33. A train running from Boston to New York meets with
an accident which causes its speed to be reduced to ,1 of what
it was before, and it is in consequence 5 hours late. If the
accident bad happened 60 miles nearer New York, the train
would have been only one hour late. "What was the rate of
the train before the accident ?
PROBLEMS. 133
34. A and B run a mile. A gives B a start of 44 yards
and beats him by 51 seconds, and afterwards gives him a start
of 1 minute 15 seconds and is beaten by 88 yards. In how
many minutes can each run a mile ?
35. A merchant has two casks, each containing a certain
quantity of wine. In order to have an equal quantity in each,
he pours out of the first cask into the second as much as the
second contained at first ; then he pours from the second into
the first as much as was left in the first; and then again from
the first into the second as much as was left in the second,
when there are found to be 1G gallons in each cask. How
many gallons did each cask contain at first ?
36. A and B arc building a fence 12G feet long; after three
hours A leaves off, and B finishes the work in 14 hours. If
seven hours had occurred before A left off, B would have fin
ished the work in 4§ hours. How many feet does each build
in one hour ?
GENERALIZATION OF PROBLEMS.
196. A problem is said to be generalized when letters are
used to represent its known quantities, as well as unknown.
The unknown quantities thus found in terms of the known
are general expressions, or formula', which may be used for
the solution of any similar problem.
197. The algebraic solution of a generalized problem dis
closes many interesting truths and useful practical rules, as
may be seen from the consideration of the following :
1. The sum of two numbers is a, and their difference is b ;
what are the two numbers ?
Let x = the greater number,
and y = the less.
By the conditions, x + y==a
x — y = b
134 ALGEBRA.
, . , . a + b . ,
►solving these equations, x = —  — , the greater number,
i a — b , ,
and y = —  — , the less.
Hence, since a and b may have any value 'whatever, the
values of x and y are general, and may be expressed as rules
for the numerical calculations in any like case; thus.
To find two numbers when their sum and difference are
given, — Add the sum and difference, and divide by 2, for the
greater of tin 1 two numbers; and subtract the difference from
the sum, and divide by 2, for the less number.
For example, if the sum of two numbers is 35, and their
difference 13,
the greater = — = 24,
and the less =  —  — = 11.
2. A can do a piece of work in a days, which it requires b
days for B to perform. In how many days can it be done if A
and B work together ?
Let x = the number of days required.
Then  = what both together can do in one day.
Also,  = what A can do in one day,
and y = what B can do in one day.
By the conditions,  +  = 
a b x
Whence, x —  — , number of days required.
Hence, to find the time for two agencies conjointly to ao
PROBLEMS. 135
complish a certain result, when the times are given in which
each separately can accomplish the same, — Divide the product
of the given tunes by their sum.
For example, if A can do a piece of work in 5 days, and B
in 4 days, the time it will take them hoth working together
•ill 5x4 20 o, ,
will be z . = — = Z% days.
5 + 4 9 J J
3. Three men, A, B, and C, enter into partnership for a
certain time. Of the capital stock, A furnishes m dollars; B,
n dollars; and C, p dollars. They gain a dollars. "What is
each man's share of the gain ?
Let x = A's share.
Then, since the shares arc proportional to the stocks,
n ./•
in
= B's share,
and == C's share.
m
ix x ty %c
By the conditions, x \ 1 — = a
in in
Whence, x = ■ , A's share.
/// + n + p
Then, = , B's share,
m m + n + p *
and — — = , C's share.
m in + n + p
Hence, to find each man's gain, when each man's stock and
the whole gain are given, — Multiply the whole gain by each
man's stock, and divide tlisproduct by the whole stock.
For example, suppose A's stock $300, B's $500, and C's
$800, and the whole gain $320.
136
ALGEBRA.
Then,
A's share
320 x 300 96000
= $60,
"300 + 500 + 800" 1600
B's share
320 x 500 160000
= $100,
"300 + 500 + 800" 1600
and C's share
320 x 800 256000
= $160.
300 + 500 + 800 1600
PROBLEMS.
4. A cistern can he rilled by three pipes ; bj the first in a
hours, by the second in b hours, and by the third in c hours.
In what time can it he filled by all the pipes running together ?
5. Using the result of the previous problem, suppose that
the first pipe fills the cistern in 2 hours, the second in 5 hours,
and the third in 10 hours. In what time can it be filled by
all the pipes running together ?
6. Divide the number a into two parts which shall have to
each other the ratio of m to n.
7. Using the result of the previous problem, divide the
number 20 into two parts which shall have to each other the
ratio of 3 to 2.
8. A courier left this place n days ago, and goes a miles
each day. He is pursued by another, starting today and
going b miles daily. How many days will the second re
quire to overtake the first ?
9. In the last example, if n = 3, a — 40, and b = 50, how
many days will he required ?
10. Required what principal, at interest at r per cent, will
amount to the sum a, in t years ?
11. Using the result of the previous problem, what principal,
at 6 per cent interest, will amount to $3108 in 8 years?
J.2. Required the number of years in which j> dollars, at r
per cent interest, will amount to a dollars.
DISCUSSION OF PROBLEMS. 137
13. Using the result of the previous problem, in how many
years will $262, at 7 per cent interest, amount to $472.91 ?
14. A banker has two hinds of money. It takes a pieces of
the first to make a dollar, and b pieces of the second to make
the same sum. If be is offered a dollar for c pieces, how many
of each kind must he give ?
15. In the last example, if a = 10, b = 20, and c= 15, how
many of each kind must he give ?
16. A gentleman, distributing some money among beggars,
found that in order to give them a cents each he should want
b cents more; he therefore gave them c cents each, and had d
cents left. Required the number of beggars.
17. A mixture is made of a pounds of coffee at m cents a
pound, h pounds at n cents, and c pounds at j> cents. Re
quired the cost per pound of the mixture.
18. A, B, and C hire a pasture together for a dollars. A
puts in m horses for t months, B puts in n horses for t' months,
and C puts in^j horses for t" months. What part of the ex
pense should each pay ?
XVII. — DISCUSSION OF PROBLEMS
LEADING TO SIMPLE EQUATIONS.
198. The Discussion of a problem, or of an equation, is the
process of attributing any reasonable values and relations to
the arbitrary quantities which enter the equation, and inter
preting the results.
199. An Arbitrary Quantity is one to which any reason
able value may be given at pleasure.
200. A Determinate Problem is one in which the given
conditions furnish the means of finding the required quantities.
138 ALGEBRA.
A determinate problem leads to as many independent equa
tions as tinre are required quantities (Art. 195).
201. An Indeterminate Problem is one in which there are
fewer imposed conditions than there are required quantities,
and, consequently, an insufficient number of independent
equations to determine definitely the values of the required
quantities.
202. An Impossible Problem is one in which the condi
tions are incompatible or contradictory, and consequently can
not be fulfilled.
203. A determinate problem, leading to a simple equation
involving only one unknown quantity, can be satisfied by but
one value of that unknown quantity (Art. 178).
An indeterminate problem, or one leading to a less number
of independent equations than it has unknown quantities, may
be satisfied by any number of values.
For example, suppose a problem involving three unknown
quantities leads to only two equations, which, on combining,
give
x — z = 10,
or, x = 10 + z.
Now, if we make z = 1, then x = 11 ;
z — 2, then x = 12 ;
z = 3, then x = 13.
Thus, we may find sets of values without limit that will sat
isfy the equation. Hence,
An indeterminate equation may have any manlier of so
lutions.
204. When a problem leads to more independent equations
than it lias unknown quantities, it is impossible.
For, suppose we have a problem furnishing three indepen
dent equations, as,
x = y+l
y — 7 — x
xy = 16
DISCUSSION OF PROBLEMS. 139
From the first two we find x = 4 and y = 3. But the third
requires their product to be 10; hence the problem is im
possible.
If, however, the third equation had not been independent,
but derived from the other two, as,
x y = 12,
then the problem would have been possible; but the last equa
tion, not being required for the solution, would have been
redundant.
INTERPRETATION OF NEGATIVE RESULTS.
205. In a Negative Result, or a result preceded by a —
sign, the negative sign is regarded as a symbol of interpre
tation.
Its significance when thus used it is now proposed to in
vestigate.
1. Let it be required to find what number must be added to
the number a that the sum may be b.
Let x = the required number.
Then, a + x = b
Whence, x = b — a.
Here, the value of x corresponds with any assigned values of
a and b. Thus, for example,
Let a = 12, and b = 25.
Then x = 25  12 = 13,
which satisfies the conditions of the problem ; for if 13 be
added to 12,* or a, the sum will be 25, or b.
But, suppose a = 30, and b = 24.
Then, x = 24  30 =  0,
140 ALGEBRA.
which indicates that, under the latter hypothesis, the problem
is impossible in an arithmetical sense, though it is possible in
the algebraic sense of the words "number," "added," and
" sum."
The negative result, — 6, points out, therefore, in the arith
metical sense, either an error or "// impossibility.
But, taking the value of x with a contrary sign, we see that
it will satisfy the enunciation of the problem, in an arithmeti
cal sense, Avhen modified so as to read :
What number must be taken from 30, that the remainder
may be 24 ?
2. Let it be required to determine the epoch at which A"s
age is twice as great as B's ; A's age at present being 35 years,
and B's 20 years.
Let us suppose the required epoch to be after the present
date.
Let x — the number of years after the present date.
Then, 35 + x = 2 (20 + x)
Whence, x = — 5, a negative result.
On recurring to the problem, we find it so worded as to
admit also of the supposition that the epoch is before the pres
ent date; and taking the value of x obtained, with the con
trary sign, we find it will satisfy that enunciation.
Hence, a negative result here indicates that a wrong choice
was made of two possible suppositions which the problem
allowed.
From the discussion of these problems we infer :
1. That negatire results indicate cither an erroneous enun
ciation of a 'problem, or a wrong supposition respecting the
quality of some quantity belonging to it.
2. That we may form, when attainable, a 'possible problem
analogous to that which involved the impassibility, or correct
DISCUSSION OF PROBLEMS. 141
the wrong supposition, by attributing to the unknown quan
tity in fin 1 equation a quality directly opposite to that
which ltml lice// attributed to it.
In general, it is not necessary to form a new equation, l»ut
simply to change in the old one the sign of each quantity
which is to have its quality changed.
Interpret the negative results obtained, and modify the
enunciation accordingly, in the following
PROBLEMS.
3. If the length of a field be 10 rods, and the breadth 8 rods,
what quantity must be added to its breadth so that the con
tents may be 60 square rods ?
4. If 1 be added to the numerator of a certain fraction, its
value becomes  ; but if 1 be added to the denominator, it be
comes §. What is the fraction ?
5. The sum of two numbers is 90, and their difference is
120 ; what are the numbers ?
6. A is 50 years old, and B 40 ; required the time when A
will be twice as old as B
7. A and B were in partnership, and A had 3 times as much
capital as B. When A had gained $ 2000. and B $ 750, A had
twice as much capital as B. What was the capital of each at
first ?
8. A man worked 14 days, his son being with him 6 days.
and received $39, besides the subsistence of himself and son
while at work. At another time lie worked 10 days, and had
his son with him 4 days, and received $28. What were the
daily wages of each '.'
142 ALGEBRA.
XVIII. — ZERO AND INFINITY.
206. A variable quantity, or simply a variable, is a quan
tity to which we may give, in the same discussion, any value
within certain limits determined by the nature of the problem ;
a constant is a quantity which remains unchanged throughout
the same discussion.
207. The limit of a variable quantity is a constant value
to which it may be brought as near as we please, but which it
can never reach.
Thus, if 3 be halved, the quotient § again halved, and so on
indefinitely, the limit to which the result may be brought as
near as we please, but which it can never reach, is zero. And,
in general, if any quantity be indefinitely diminished by di
vision, its limiting value is zero.
208. If any quantity be indefinitely increased by multipli
cation or otherwise, its limiting value is called Infinity, and is
denoted by the symbol co .
209. It is evident, from the definition of Art. 207, that if
two variable quantities are always equal, their limiting values
will be equal.
210. We will now show how to interpret certain forms
which may be obtained in the course of mathematical opera
tions.
. ' a a
Let us consider the fraction  : and let  = a*.
b b
1. Interpretation of
Let the numerator of remain constant, and the denomi
h
nator be indefinitely diminished by division. By Art. l'>7. it'
the denominator is divided by any quantity, the value of the
ZERO AND INFINITY. 143
fraction is multiplied by that quantity ; hence the value of the
fraction, x, increases indefinitely as b is diminished indefi
nitely. The limiting value of b being (Art. 207), the limit
ing value of  will be  ; and the limiting value of x is co
b a
(Art. 208). Now  and x being two variable quantities always
equal, by xlrt. 209 their limiting values are equal ; or,
a
a
2. Interpretation of — .
00
Let the numerator remain constant, and the denominator be
indefinitely increased by multiplication. By Art. lo8, if the
denominator is multiplied by any quantity, the value of the
fraction is divided by that quantity ; hence x is diminished
indefinitely by division as the denominator increases in
definitely. The limiting value of b being oo, the limiting
ft ct
value of  will be — : and the limiting value of x is 0. By
b co
Art. 209 these limiting values are equal ; or,
GO
Problem of the Couriers.
211. The discussion of the following problem, commonly
known as that of Clairaut, will serve to further illustrate
the form , besides furnishing us with an interpretation of the
,
form pr .
Two couriers, A and B, are travelling along the same road,
in the same direction, \V R, at the rates of m and n miles per
hour respectively. If at any time, say 12 o'clock, A is at the
144 ALGEBKA.
point P, and B a miles from him at Q, when and where are
they together '.'
I ! 1 1
R' P Q B
Let #= the required time in hours,
and x = the distance A travels in the time t, or the dis
tance from P to the place of meeting.
Then x — a = the distance B travels in the time t, or the dis
tance from Q to the place of meeting.
Since the distance equals the rate multiplied by the time,
x = m t
x — a ==n 1
Solving these equations with reference to t and x,
a
x =
It is proposed now to discuss these values on different sup
positions.
1. in > n.
This hypothesis makes the denominator m — n positive;
hence the values of both t and x are positive. That is, the
couriers are together after 12 o'clock, and to the right of P.
This interpretation corresponds with the supposition made.
For, if A travels faster than B, he will eventually overtake
him, and in advance of their positions at 12 o'clock.
2. in < n.
This hypothesis makes the denominator m — n negative;
hence the values of both t and x are negative. Now. from
what we have observed in regard to negative results < Art. 205),
these values of t and x indicate that the couriers w< re together
before 12 o'clock, and to the left of P.
111 
 n
m
a
m
 n
ZERO AND INFINITY. 145
This interpretation corresponds with the supposition made.
For, if A travels more slowly than B, he will never overtake
him ; but as they are travelling along the same road, they
must have been together before 12 o'clock, and before they
could have advanced as far as P.
3. m = n.
This hypothesis makes the denominator m — n equal to zero ;
ft 77h CL
so that the values of t and x become  and j—, respectively;
or, by Art. 210, t = oo and x — co . Since from its nature
(Art. 208), go is a value which we can never reach, the values
of t and x may be regarded as indicating that the problem is
impossible under the assumed hypothesis.
This interpretation corresponds with the supposition made.
For, if the couriers were a miles apart at 12 o'clock, and were
travelling at the same rate, they never had been and never
would be together.
Thus, infinite results Indicate the imjjossibility of a problem.
4. a = 0, and m > n or m < n.
By this hypothesis, the values of t and x each become
m — n '
or (Art. 102), £ = and x = 0. That is, the couriers are to
gether at 12 o'clock, at the point P, and at no other time and
place.
This interpretation corresponds with the supposition made;
for, if the distance between them at 12 o'clock is nothing, they
are together at P ; but as their rates are unequal, they cannot
be together after 12 o'clock, nor could they have been together
before that time.
5. a = 0, and m = n.
By this hypothesis, the values of t and x each take the
form^.
146 ALGEBRA.
Referring to the enunciation of the problem, we see that if
the couriers were together at 12 o'clock, and were travelling
at the same rate, they always had been, and always would be,
together. There is, then, no single answer, or finite number
of answers, to the problem in this case; and results of this
form are therefore called indeterminate.
Thus, a result  indicates indeterminathn.
212. The symbol , however, does not always represent an
indeterminate quantity which may have any fin ite mine. Now,
in the preceding problem the result  was obtained in conse
quence of two independent suppositions, one causing the nu
merator to become zero, and the other the denominator. We
say independent^ because the quantity m — n can be equal to
without necessarily causing the quantity a to become 0. And
in all similar cases, we should find the result  susceptible of
the same interpretation.
But if the symbol  is obtained in consequence of the same
supposition causing both numerator and denominator to be
come zero, it will be found to have a single definite limiting
value.
a 2 — b
Take, for example, the fraction — ; if b = a, this single
supposition causes both numerator and denominator to become
zero, and the fraction takes the form .
Now, dividing both terms by a — b, we have
o?V _ a + b
a*ab~ a ' { '
which equation is true so Ion*; as b is not equal to a. It is
not necessarily true when b is equal to a, because the second
ZERO AND INFINITY. 147
member was obtained by dividing both terms of the first mem
ber by a — h (which divisor becomes when b = a), as we
cannot speak of dividing a quantity by nothing.
In (1), as b approaches a, the limiting value of the first
member is , and the limiting value of the second member
is 2. Thus we have (Art. 209), Q = 2.
Hence the limiting value of the fraction, as b approaches a,
is 2.
213. A proper understanding of the theory of indetermi
nation, and of the relation of zero to finite quantities, will lead
to the detection of the fallacy in some apparently remarkable
results.
For example, let a = b
Then a i = a b
Subtracting //, a 2 — J 2 = a b — b' 2
Factoring, (a + b) (a — b)—b (a — b) (1)
Dividing by a — b, a + b = b (2)
But b = a; hence a + a = a
then 2 a ■ = a
or, 2=1
The error was made in passing from (1) to (2). Equation
(1) may be written
a + b a — b
b a — b
Now, as b = a, the second member is an expression of the* form
 . But we assumed in going from (1) to (2) that  —  = 1,
" ' Cv " (J
or that  = 1 ; which we have seen in Arts. 211 and 212 is not
necessarily the case, as it may have any value whatever.
148 ALGEBRA.
XIX. — INEQUALITIES.
214. An Inequality is an expression indicating that one of
two quantities is greater or less than the other ; as,
a > b, and m < n.
The quantity on the left of the sign is called the first mem
ber, and that on the right, the second member of the inequality.
215. Two inequalities are said to subsist in the same sense
when the first member is the greater or less in both.
Thus,
a > b, and c > d ; or 3 < 4, and 2 < 3,
are inequalities which subsist in the same sense.
216. Two inequalities are said to subsist in a contrary
sense, when the first member is the greater in the one, and
the second in the other. Thus,
a > b, and c < d ; or x < y, and u > z,
are inequalities which subsist in a contrary sense.
217. In the discussion of inequalities, the terms greater and
less must be taken as having an algebraic meaning. That is,
Of" 11 !/ two quantities, a and b, a is the greater when a — b
is positive, and a is the less when a — b is negative.
Hence, a negative quantity must be considered as less than
nothing; and, of two negative quantities, that is the greater
which has the least number of units (Art. 49). Thus,
>  2, and  2 >  3.
218. An inequality will continue in the same sense after
the same quantity has been added to, or subtracted from, each
member.
INEQUALITIES. 149
For, suppose a > b ;
then, by Art. 217, a — b is positive ; consequently,
(a + c) — (b + c) and (« — e) — (ft — c)
are positive, since each equals a — b. Therefore,
a + c > b + c, and a — c > S — c.
Hence, it follows that a term may be transposed from one
member of an inequality to the other, if its sign be changed.
219. If the signs of all the terms of an inequality be
changed, the sign of inequality must be reversed.
For, to change all the signs, is equivalent to transposing
each term of the first member to the second, and each term of
the second member to the first.
220. If two or more inequalities, subsisting in the same
sense, be added, member to member, the resulting inequality
will also subsist in the same sense.
For, let
a> b, a'> V, a"> b",
then, by Art. 217, a — b, a' — b 1 , a" — b", are all positive ;
and consequently their sum
a + a' + a" + —b — b' — b"—
or, (a + a' + a"+ )  (b + V + b" + )
i> positive. Hence,
a + a'+ a!'+ > b + b' + b" +
221. If two inequalities, subsisting in the same sense, be
subtracted, member from member, the resulting inequality will
not always subsist in the same sense.
150 ALGEBRA.
For, let
a > b, and a' > V ;
*
■
then a — b and a! — b 1 are positive ; but a — b— (a' — b'), or
{a — a') — (b — &')> ma y l )e either positive, negative, or 0.
That is,
a — a'> b — b', a — a' < & — b', or a — a' = b — V.
222. '// inequality will continue in the same sense after
each member has been multiplied or divided by the same posi
tive quantity.
For, suppose a > 5 ;
then, since « — J is positive, if m is positive,
vi (a — b) and — (a — b)
m K '
are positive. That is, m a— m b and are positive.
in in
Hence,
7 i a b
m a > ??i y, and — > — .
in m
223. If each member of an inequality he multiplied or di
vided by the same negative quantity, the sign of inequality
must be reversed.
For, since multiplying or dividing by a negative quantity
must change the signs of all the terms, the sign of inequality
must be reversed (Art. 219).
224. The solution of an inequality consists in determining
the limit in the value of its unknown quantity.
This may be done by the application of the preceding prin
ciples.
When, however, an inequality and an equation are given,
containing two unknown quantities, the process of elimination
will be required in the solution.
INEQUALITIES. 151
In verifying an inequality, if the symbols of the unknown
quantities be taken equal to their respective limits, the ine
quality becomes an equation.
EXAMPLES.
225. 1. Find the limit of x in the inequality
23 2x „
i x  j > g + 5.
Clearing of fractions, 21 a; — 23 > 2 x + 15
Transposing, and uniting, 19 ic > 38
Whence, x > 2, ^4«s.
2. Find the limits of x in the inequalities,
ax + 55x5«i >a 2 (1)
5x7ftj; + 7«Ki' 2 (2)
from (1), ax + 5b x > a 2 + 5 ab
x (a + 5 b) > a (a + 5 b)
x > a.
From (2), bx — 7ax<b 2 — 7ab
x (b  7 a) < b (b  7 a)
x< b.
Hence, x is greater than a, and less than b, Ant;.
3. Find the limits of x and y in the following inequality and
equation :
4 x + 6 y > 52 (1)
4 jc + 2 y = 32 (2)
Subtracting (2) from (1), 4 ?/ > 20
2/ > 5. (3)
From (2), we have y = 16 — 2 x
152 ALGEBRA.
Substituting in (3), 16  2 x > 5
2x >ll
11
*> 2
or (Art. 219),
*<T
Hence,
y > 5, and #
4. Given 5 a? — 6 > 19. Find the limit of cc.
5. Given 2x5 >25; 3x7<2x+13. Find tin
limits of x.
6. Given 3 z + 1 > 13 — x; 4cc — 7 < 2 a; + 3. Find an
integral value of x.
7. Given 5# + 3?/>46 — y; ?/ — z = — 4. Find the lim
its of x and y.
ex c d x cl
8. Given — + dx — crf> T ; 5 c sc + c d < ^. Find
the limits of x.
9. Given 2x + 3y<57; 2x + y = 32. Find the limits
of x and y.
10. A teacher being asked the number of his pupils, replied
that twice their number diminished by 7 was greater than 29 ;
and that three times their number diminished by 5 was less
than twice their number increased by 16. Required the num
ber of his pupils.
11. Three times a certain number, plus 16, is greater than
twice that number, plus 24 ; and two fifths of the number, plus
5, is less than 11. Required the number.
12. A shepherd lias a number of slice]) such that three times
the number, increased by 2, exceeds twice the number, in
creased by 61; and 5 times the number, diminished by 70, is
less than 4 times the number, diminished by 9. How nianv
sheep has he ?
INVOLUTION. 153
XX.  INVOLUTION.
226. Involution is the process of raising a quantity to any
required power.
This may he effected, as is evident from the definition of a
power (Art. 17), by taking the given quantity as a factor as
many times as there are units in the exponent of the required
power.
227. If the quantity to he involved is positive, the signs of
all its powers will evidently be positive ; but if the quantity is
negative, all its even powers will be positive, and all its odd
powers negative. Thus,
(— a f =(a)x (— a) X (— a) = + cr X (— a) = — a 3 ,
( a y = ( a) X ( a) X ( a) x ( a) = ( a 3 ) X ( a) = + a\
and so on.
Hence,
Every even poiver is positive, and every odd power has the
same sign as its root.
INVOLUTION OF MONOMIALS.
228. 1. Let it be required to raise 5 a" b c 3 to the fourth
power.
5 a 2 bc 3 x5a 2 bc s x5aHc 3 x5a 2 b c 3 = 625 a 8 b* c v2 , Ans.
2. Raise —3m n 3 to the third power.
(— 3 m n 3 ) X (— 3 m n 3 ) X (— 3 m n 3 ) = — 21 m 3 ?j 9 , Ans.
RULE.
Raise the numerical coefficient to the required power, and
multiply the exponent of each letter by the exponent of the re
(j uired power ; making the sign of every wen poiver positive,
and the sign of every odd power the same as that of its root.
154 ALGEBRA.
EXAMPLES.
j
Find the values of the following :
3. (a 2 x) 2 . 7. (2x m )\ 11. (2ab n xf.
4. (3 cr b) s . 8. (2ab 2 x 8 ) 5 . 12. (7w*ra) 4 .
5. (ab 2 c 3 y. 9. (a 2 b 2 )\ 13. (5a 2 6 8 c 4 ) 8 .
6. (a n b) m . 10. (a 2 c 3 ) 3 . 14. (6 a; 3 ?/ 7 ) 3 .
A fraction is raised to any required power by raising both
numerator and denominator to the required power.
Thus,
2x*\*_{ 2x*\ ( 2x 2 \ f 2x*\ 8 a; 6
'3f) ~\ 3y) X \ 3y 3 l X \ 3y 3 )~~ 21 if
Find the values of the following :
15 . m\ it. L^r. ia i 2x ^ z
b J * V 3b I ' V 36
16. R**\\ 18. fL'^V. 20. f*'^'
4 x y*/ \o / V 4 a~
INVOLUTION OF POLYNOMIALS.
229. Polynomials may he raised to any power, as is obvious
from Art. 226, by the process of successive multiplications.
Thus,
(a + b) 2 =(a + b)(a + b) = a 2 + 2ab + b' 2 ,
(a + b) 3 =(a + b) (a + b) (a + b) = a 3 + 3 a 2 b + 3 a b 2 + b 3 ,
and so on. Hence the following
RULE.
Multiply the polynomial by itself, until it has been taken as
a factor as many times as there are units in the exponent of
the required power.
INVOLUTION. 155
EXAMPLES.
Find the values of the following :
1, (a b) s . 3. {l + a 2 + b 2 )\ 5. (a m a n )\
2. (^) 2  4. (a + mnf. 6. (a + b)
5
In Chapter XXXVII will be given a method for raising a
binomial to any required power, without going through with
the process of actual multiplication.
SQUARE OF A POLYNOMIAL.
230. It has been shown (Arts. 104 and 105) that the
square of any binomial expression can be written down, with
out recourse to formal multiplication, by application of the
formulae
(a + b) 2 = a 2 + 2ab + b 2 ,
(aby = a?2ab + b 2 .
We may also show, by actual multiplication, that
(a + b + c) 2 =a 2 + 2ab + 2ac + b 2 + 2bc + c 2 ,
(a + b + c + d) 2 = a 2 + 2 ab + 2 ac + 2 ad + b 2 + 2 bc+ 2bd
+ c 2 + 2cd + d 2 ,
and so on.
These residts, for convenience of enunciation, may be writ
ten in another form,
(a + b) 2 = a 2 + b 2 + 2ab,
(a — b) 2 = a 2 + b 2 — 2 a b,
(a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2 a c + 2b c,
(a + b + c + d) 2 = a 2 + b 2 +c 2 +d 2 +2ab + 2ac + 2ad
+ 2bc + 2bd + 2cd,
and so on. Hence, the following
156 ALGEBRA.
RULE.
Write the square of each term, together with twice its prod
uct by each of the terms following it.
1. Square x 2 — 2 x — 3.
Square of each term, a? 4 + 4 x 2 +9
Twice x 2 X the terms following, — 4. x 3 — 6x 2
Twice — 2 x X the term following, + 12 x
»
Adding, the result is x A — 4:X 3 — 2x 2 +12x + 9.
EXAMPLES.
Square the following expressions :
2. a — b + c. 8. 1 + x + x 2 + x 3 .
3. 2x 2 + 3x + 4. 9. x 3 4x 2 2x3.
4. 2x 2  3x + i 10. 2x 3 +x 2 +l xl.
5. a — b — c + d. l\. x 3 + bx 2 — x + 2.
6. »i 3 + 2a; 2 + a;+2. 12. 3x 3 2 x 2 x+ ±.
7. 1 — 2 a; + 3 ar. 13. a + & — c — d + e.
CUBE OF A BINOMIAL.
231. Hy actual multiplication we may show,
(a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 ,
(ab) 3 = a 3 3a 2 b + 3a b 2  b 3 .
Hence, for finding the cube of a binomial, the following
RULE.
Write the cube of the first term, phis three times the square
of the first term times the second, 'plus three times the first
term times the square of the second, plus the cube of the second
term.
3\3
INVOLUTION. 157
EXAMPLES.
1. Find the cube of 2 x 2  3 y 3 .
(2 a 2 ) 3 + 3 (2 x 2 ) 2 ( 3 y 3 ) + 3 (2 a: 2 ) ( 3 y 3 ) 2 + ( 3 if)
' = 8z 6 + 3 (4* 4 ) (3t/ 3 ) + 3 (2 a; 2 ) (9 if) + (27 y 9 )
_ 8 x G  36 x* f + 54 x 2 if  27 y 9 , Jns.
Cube the following :
2. a 2 +2b. 4. 3 a; 4. 6. 4 a; 2 ay.
3. 2m + 5«. 5. 2z 3 3. 7. 3a;7/ + 5a6
.2
CUBE OF A POLYNOMIAL.
232. By actual multiplication we may show,
(a + ft + e y = a 3 + b 3 + c 3 + 3 a 2 b + 3 a 2 c + 3 b 2 a + 3 b 2 c
+ 3 c 2 a + 3c 2 + Gabc,
( a + l ) + c + d) 3 = a 3 + b 3 + c 3 + d 3 + 3a 2 b + 3a 2 c + 3a 2 d
+ 3b 2 a + 3 b 2 c + 3b 2 d + 3 c 2 a + 3 c 2 6
+ 3 c 2 d + 3 d 2 a + 3d 2 b + 3d 2 c+6abc
+ 6 a b d + 6 a c d + 6 b c d,
and so on. Hence, for finding the cube of a polynomial, the
following
BULE.
Write the cube of each term, tor/ether with three times the
product of its square by each of the other terms, and also six
times the product of every three different terms.
EXAMPLES.
1. Find the cube of 2 a; 2 .— 3 x — 1.
8 x 6 21 x 3  1
 36 x h  12 a 4
f 54 a; 4  27 x 2
+ Qx 2 9x
+ 36x 3
8 x 6  36 x s + 42 x 4 + 9 a' 3  21 x 2  9 x  1, Arts.
158 ALGEBRA.
Find the cubes of the following :
2. a + b — c. 5. 2  2 .r + x\
3. a; 2 21. 6. 1 + a; + x 2 + X 3 .
4. a 6 + 1. 7. 2.« 3 r + 2x3.
XXI. — EVOLUTION.
233. Evolution is the process of extracting any required
root of a quantity.
This may be effected, as is evident from the definition of a
root (Art. 17), by determining a quantity which, when raised
to the proposed power, will produce the given quantity. It is,
therefore, the reverse of involution.
234. Any quantity whose root can be extracted is called a
perfect power ; and any quantity whose root cannot be ex
tracted is called an imperfect power.
A quantity may be a perfect power of one degree, and not of
another. Thus, 8 is a perfect cube, but not a perfect square.
235. To extract any root of a simple quantity, the expo
nent of that quantity must be divided by the index of the root.
For, since the ?ith power of a m is a mn (Art. 228), it follows
that the nth. root of a mn is a m .
236. Any root of the product of two or more factors is
equal to the product of the same root of each of the factors.
For, we have seen in Art. 228, in raising a quantity com
posed of factors to any required power, that cadi of the factors
is raised to the same power.
237. From the relation of a root to its corresponding
power, it follows, from Art. 227, that
EVOLUTION. 159
1. The odd roots of any quantity have the same sign as the
quantity.
Thus, \j a 3 = a ; and ^ — a 5 = — a.
2. The even roots of a 'positive quantity are either positive
or negative.
For either a positive or negative quantity raised to an even
power is positive. Thus,
y/ a 4 = a or — a ; or, y' a* = ± a.
Note. The sign ±, called the double sign, is prefixed to a quantity
■when we wish to indicate that it is either + or  .
3. Even roots of a negative quantity are not possible.
For no quantity raised to an even power can produce a neg
ative result. Such roots are called impossible or imaginary
quantities.
EVOLUTION OF MONOMIALS.
238. From the principles contained in Arts. 235 to 237,
we obtain the following
RULE.
Extract the required root of the numerical coefficient, ami
divide the exponent of each letter by the index of the root;
making the sign of every even root of a positive quantity ±,
and the sign of every odd root of any quantity the same as
that of the quantity.
If the given quantity is a fraction, it follows from Art. 228
that we may fake the required root of both of its terms.
EXAMPLES.
1. Find the square root of 9 a 4 b c 6 .
\/9a 4 i 2 c 6 = ±3fl 2 i c s , Ans.
2. Find the cube root of  64 a 9 x z y 6 .
.3/
^ _ 64 a 9 x s y G = — 4:a 3 x y 2 , Ans.
160 ALGEBRA.
8 X 3 7)1
3. Find the cube root of
27 a G b 9
8 / /8 x 3 m 12 \ _ 2 x m 4
Find the values of the following :
*
4. if  125 x 3 y 6 . 9. yV" 1 ^. 14. \/ 729 a 1 * b 24 <A
5. ySla*b 8 . 10. Sl8an«x\ 15. \/32 c 6 "^ ™.
_ J f 32 m 5 n 10 \ . 5/
V \ 243 J ' 1L ^ 16 a;2m+2 a2 " 16  V 243 w 16 »*■
7. \/l»^?. 12. y/ ^QQ^l ) • 17. V(« + *)W
8. y/ 625 a 12 c 2 . 13. y' 3 2 " 6 3n a n . 18. faj 8B + V 1 " 6 
SQUARE ROOT OF POLYNOMIALS.
239. In Art. 11G we explained a method of extracting the
square root of a trinomial, provided it was a perfect square.
We will now give a method of extracting the square root of
any polynomial which is an exact square.
Since the square of a + b is a 2 + 2 a b + b 2 , we know that
the square root of a 2 + 2 a b + b 2 is a + b. If we can discover
an operation by which we can derive a + b from a 2 + 2 a b + b 2 ,
we can give a rule for the extraction of the square root.
„ „ 7 7 „ 7 Arranmne the terms of the
a 2 + 2ab+b 2 a + b 5 °. '
2 square according to the descend
o „ , 7. 2 a b A V 1 * n 8 powers of a, we observe thai
2 a b + b 2 the square root of the first term,
a 2 , is a, which is the first term
of the required root. Subtract its square, a 2 , from the uiven
polynomial, and bring down the remainder, 2 a b + b 2 or
(2 a + b) b. Dividing the first term of the remainder by 2 a,
that is, by twice the first term of the root, we obtain b, the
other term. This, added to 2 a, completes the divisor, 2 a + b ;
EVOLUTION. 161
which, multiplied by b, and the product, 2 ab + V 2 , subtracted
from the remainder, completes the operation.
By a similar process, a root consisting of more than two
terms may be found from its square. Thus, by Art. 230, we
know that (a + b + c) 2 = a 2 + 2 a b + b 2 + 2 a c + 2 b c + c 2 .
Hence, the square root of a 2 + 2 a b + b' 2 + 2 a c + 2 b c + c 2 is
a + b + c.
a 2 + 2ab + b 2 + 2ac+2bc + c 2
•2
a
a + b + c
2a + b
2ab + b 2 + 2ac+2bc + c 2
2ab + b 2
2 a + 2 b + c
2ac+2bc + c 2
2ac+2be + r 2
The square root of the first term, a 2 , is a, which is the first
term of the required root. Subtracting a 2 from the given poly
nomial, we obtain 2 a b as the first term of the remainder.
Dividing this by twice the first term of the root, 2 a, we ob
tain the second term of the root, b, which, added to 2 a, com
pletes the divisor, 2 a + b. Multiplying this divisor by b, and
subtracting the product, 2 a b + b' 2 , from the first remainder,
we obtain 2 a c as the first term of the next remainder.
Doubling the root already found, giving 2 a + 2 b, and di
viding the first term of the second remainder, 2 a c, by the first
term of the result, 2 a, we obtain the last term of the root, c.
This, added to 2 a + 2 b, completes the divisor, 2 a + 2 b + c ;
which, multiplied by the last term of the root, c, and subtracted
from the second remainder, leaves no remainder.
From these operations we derive the following
RULE.
Arrange the terms according to the powers of some letter.
Find the square root of the first term, write it as the first
term of the root, and subtract its square from the given poly
nomial.
Divide the first term of the remainder by double the root
already found, and add the result to the root, and also to the
divisor.
162
ALGEBRA.
Multiply the divisor as it now stands by the term of the root
last obtained, and subtract the product from the remainder.
If there are other terms remaining, continue the operation
In the same manner as before.
Note. Since all even roots have the double sign ± (Art. 237), all the
terms of the result may have their signs changed. In the examples, how
ever, we shall consider only the positive sign of the result.
EXAMPLES.
1. Find the square root of 9 x* — 12 x s + 16 x 2 — 8 x + 4.
9a; 4 12a; 3 + lGx~8x + 4: 3a; 2 2a; + 2
9 a; 4
6 a: 2  2
./'
12 a; 3
 12 x 3
4 a;' 2
6 x  4 x + 2
12 x 2  8 x + 4
12 a;' 2 . 8 x + 4
Ans. 3x 2 — 2x + 2.
Find the square roots of the following :
2. ±x i ±x s 3x 2 + 2x + l.
2 1
— + — i
m m
3. 4 a 4 16 a 3 +24 a 2  16 a + 4.
4. m 2 + 2 m — 1
5. 9 — 12 x + 10 :•■  4 x 3 + x*.
6. 19 x 2 + 6 x 3 + 25 + x i + 30 a.
7. 28 a 3 + 4 a; 4  14 x + 1 + 45 x 2 .
8. 40 jc + 25  14 x 2 + 9 x*  24 a; 3 .
9. 4 .r 4 + 64  20 a 3  80 x + 57 x 2 .
10. a 2 + b 2 + c 2 2 ab 2 a c + 2 b c
11. a; 2 + 4 y 2 + 9 « 2 — 4 a; ?/ + 6 a; » — 12 y z.
No rational binomial is an exact square;; hut, by the rule,
the (ip/imxliiinte root may be found.
EVOLUTION. 10
• i
Find, to four terms, the approximate square roots of the fol
lowing :
12. 1 + x. 13. a 2 + b. 14. 1 — 2 x. 15. a 2 + x 2 .
The square root of a perfect trinomial square may be ob
tained by the rule of Art. 116,
Find the square roots of the first and last terms, and con
nect the results by the sign of the second term.
Extract the square roots of the following :
16. x 4 + 8x 2 +16. 19. « 2m 4 a m+n + 4 a 2 ".
rtrt a 2 4 a ±b 2
17. 9x*6xf + f. 20. _ — +^.
t 2 4 x 2 9 ?/ 4
18. a*ax + T . 8L 9? + 2 +4^
SQUARE ROOT OF NUMBERS.
240. The method of Art. 239 may be used to extract the
square roots of numbers.
The square root of 100 is 10 ; of 10000 is 100 ; of 1000000,
is 1000 ; and so on. Hence, the square root of a number less
than 100 is less than 10 ; the square root of a number between
10000 and 100 is between 100 and 10 ; the square root of a
number between 1000000 and 10000 is between 1000 and 100 ;
and so on.
Or, in other words, the integral part of the square root of a
number of one or two figures, contains one figure; of a number
of three or four figures, contains two figures ; of a number of
five or six figures, contains three figures ; and so on. Hence,
If a point is placed over evei*y second figure in any integral
number, beginning with the units' 1 place, the number of point*
will shoiv the number of figures in the integral part of its
square root.
164 ALGEBRA.
241. Let it be required to find the square root of 4356.
Pointing the number according to
60+6 the rule of Art. 240, it appears that
there are two figures in the integral
4356
3600
120 + 6 756 'J" ° ~~— ~ .
jgQ part oi the square root. Let a denote
the figure in the tens' place in the
root, and b that in the units' place. Then a must be the
greatest multiple of 10 whose square is less than 4356 ; this
we find to be 60. Subtracting a 2 , that is, the square of 60, or
3600, from the given number, we have the remainder 756.
Dividing this remainder by 2 a, or 120, gives 6, which is the
value of b. Adding this to 120, multiplying the result by 6,
and subtracting the product, 756, there is no remainder.
Therefore we conclude that 60 + 6, or 66, is the required square
root.
The zeros being omitted for the sake of brevity, we may ar
range the work in the following form :
4356
36
G6
126
756
756
RULE.
Separate the given number into periods, by pointing every
second figure, beginning with the units' [dace.
Find the greatest square in the lefthand period, and place
its root on the right ; subtract the square of this root from the
first period, and to the remainder bring down the next period
for a dlr hi end.
Divide this dividend, omitting the last figure, by double the
root already found, and annex the result to the root and also
to the divisor,
Multiply tin' divisorj as it now stands, by the figure of the
root last obtained, and subtract the product from tin dividend.
If there are more periods to be brought down, continue the
operation in the same manner as before.
EVOLUTION. 165
If there be a final remainder, the given number has not an
exact square root ; and, since the rule applies equally to deci
mals, we may continue the operation, by annexing periods of
zeros to the given number, and thus obtain a decimal part to
be added to the integral part already found.
It will be observed that decimals require to be pointed to
the right ; and if they have no exact root, we may continue
to form periods of zeros, and obtain decimal figures in the root
to any desirable extent.
As the trial divisor is necessarily an incomplete divisor, it is
sometimes found that after completion it gives a product' larger
than the dividend. In such a case, the last root figure is too
large, and one less must be substituted for it.
The root of a common fraction may be obtained, as in Art.
238, by taking the root of both numerator and denominator,
when they are perfect squares. If the denominator only is a
perfect square, take the approximate square root of the nu
merator, and divide it by the square root of the denominator.
If the denominator is not a perfect square, either reduce the
fraction to an equivalent fraction whose denominator is a per
fect square, or reduce the fraction to a decimal.
EXAMPLES.
1. Extract the square root of 49.434961.
49.434961
49
7.031
1403
4349
4209
14061
14061
14061
Ans. 7.031.
Here it will be observed that, in consequence of the zero in
the root, we annex one zero to the trial divisor, 14, and bring
down to the corresponding dividend another period.
Extract the square roots of the following :
ALGEBRA.
6.
.9409.
10.
.006889.
7.
6561
9025'
11.
.0000107584
8.
1.170724.
12.
811440.64.
9.
446.0544.
13.
.17015625.
166
2. 273529.
3. 45796.
4. 106929.
5. 33.1776.
Extract the square roots of the following to the fifth decimal
place :
14. 2. 16. 31. 18.
15. 5. 17. 173. 19.
242. When n + 1 figures of a square root have been ob
tained by the ordinary method, n more may be obtained by
simple division only, supposing 2n + l to be the whole num
ber.
Let N represent the number whose square root is required,
a the part of the root already obtained, x the rest of the root ;
then
y/_ZV= a + x,
whence, iV= a 2 + 2 a x + x 2 ;
therefore, JV — a 2 = 2 a x + x 2 ,
7
1
20.
9'
3'
3
2
21.
16'
Na*
= x +
x~
2 a '2a
Then iV— a 2 divided by 2 a will give the rest of the square
x
root required, or x, increased by = — ; and we shall show that
x
= — is a proper fraction, less than \, so that by neglecting the
Zi a
remainder arising from the division, we obtain the part re
quired. For, x by supposition contains n figures, so that x 2
cannot contain more than 2 n figures ; but a contains 2n + l
EVOLUTION.
167
figures ; and hence — is a proper fraction. Therefore 
a 2 a
is a proper fraction, and less than ^.
In the demonstration we supposed JV an integer with an
exact square root ; but the result may be extended to other
cases.
From the examples in Art. 241, we observe that each re
mainder brought down is the given expression minus the
square of the root already obtained ; and is therefore in the
form A 7 " — a 2 . If, then, any remainder be divided by twice
the root already found, we can obtain by the division as many
more figures of the root as we already have, less one.
We will apply these principles to calculating the square root
of 12 to the. sixth decimal place. We will obtain the first four
figures of the result by the ordinary method :
12.000000
9
3.464
64
300
256
686
4400
4116
6024
28400
27696
'04
The remainder now is .000704 ; and twice the root already
found is 6.928. Then, by dividing .000704 by 6.928, we can
obtain the next three figures of the root. Thus,
6.928). 0007040 (.000102
.0006928
11200
That is, the square root of 12 to the nearest sixth decimal
place is 3.464102.
The following rule will be found to save trouble in obtaining
approximate square roots by this method :
168
ALGEBRA.
Divide the remainder by twice the root already found {omit
ting the decimal point), and annex all of the quotient, except
the decimal point, to the part of the root already found.
In practice the work would be arranged thus :
12.
9
64 30C
25C
3.464
)
686
4400
4116
6924
: 28400
27696
6928) 704.000 (.102
6928
11200
Am. 3.464102
EXAMPLES.
1. Extract the square root of 11 to the 4th decimal place.
2. Extract the square root of 3 to the 6th decimal place.
3. Extract the square root of 61 to the 8th decimal place.
4. Extract the square root of 131 to the 3d decimal place.
5. Extract the square root of 781 to the 5th decimal place.
6. Extract the square root of 12933 to the 4th decimal place.
CUBE ROOT OF POLYNOMIALS.
243. Since (a + b) 3 = a s + 3 a 2 b + 3 a b 2 + b 3 , we know
that the cube root of a 3 + 3 a" b + 3 a b 2 + b 3 is a + b.
a 3 + 3 a 2 b + 3 a b 2 + b 3
a"
a + b
3a 2 +3ab + b 2 3 a 2 b + 3 a b 2 + b*
3a 2 b + 3ab 2 + b 5
EVOLUTION. 1G9
Arranging the terms of the cube according to the descending
powers of a, we observe that the cube root of the first term, a 3 ,
is a, which is the first term of the required root. Subtract its
cube, a 3 , from the given polynomial, and bring down the re
mainder, 3 a 2 b + 3 a b' 2 + b 3 or (3 a 2 + 3 a b + b' 2 ) b. Dividing
the first term of the remainder by 3 a 2 , that is, by three times
the square of the first term of the root, we obtain b, the other
term of the root. Adding to the trial divisor 3 a b, that is,
three times the product of the first term of the root by the last,
and b' 2 , that is, the square of the last term of the root, completes
the divisor, 3 a' 2 + 3 a b + b' 2 ; which, multiplied by b, and the
product, 3 a 2 b + 3 a b' 2 + b 3 , subtracted from the remainder,
completes the operation.
If there were more terms, we should proceed with a + b
exactly as previously with a ; regarding it as one term, and
dividing the first term of the remainder by three times its
square ; and so on. Hence, the following
RULE.
Arrange the terms according to the powers of some letter.
Find the cube root of the first term, write It as the first term
of the root, and subtract its cube from the given polynomial.
Take three times the square of the root already found for a
trial divisor, divide the first term of the remainder by it, and
write the quotient for the next term of the root.
Add to the trial divisor three times the prod net of the first
term by the second, and the square of the second term.
Multiply the complete divisor by the second term of the root,
and subtract the product from the remainder.
If there are other terms remaining, consider the root already
found as one term, and proceed as before.
EXAMPLES.
1. Find the cube root of x 6 — 6 x'° + 40 x 3 — 96 x — 64.
170
X
X
ALGEBRA.
6 6a 5 +40a; 3 96a64 1 x 2 2j
6
3x i 6x a +4:x i
—6 a 5
— 6 a; 5 +12 a 4  8 x s
3 a 4  12 a 3 + 12 a; 2
12a 2 +24a; + 16
12a; 4 +48a; 3
3a 4 12a: 3 + 24a
+ 16
 12 x* +48 x 3  96 .r 64
Ans. x' 2 — 2 a; — 4.
The formation of the second divisor may be explained thus :
Regarding the root already obtained, a 2 — 2 a, as one term,
three times its square gives 3 a 4 — 12 a 3 + 12 x' 2 ; three times
x~ — 2 a; times — 4 ; gives — 12 x' 2 + 24 x ; and the square of the
last root term is 16. Adding these results, we have for the
complete divisor, 3 a 4 — 12 x 3 + 24 x + 16.
Find the cube roots of the following:
2. 1 — 6 y + 12 y' 2  8 ?/ 3 .
3. 8 a; 6 + 36 a; 4 + 54 a; 2 + 27.
4. 64 a; 3  144 a b x 2 + 108 a 2 i 2 a:  27 a 3 6 3 .
5. a 6 + 6 a 5  40 a 3 + 96 x — 64.
6. v/«l + 57/ 3 37/ 5 3y.
7. a: 3 + 3a;H hg.
8. 15 r 4  6 a  6 a; 5 + 15 x 2 + 1 + a 6  20 x*.
9. a a + 3 a a & + 3a 2 c + 3aJ 2 + 6a6c + 3ac 2 +S 8 + 3 6 a c
+ 3l> r 2 + c 8 .
10. 9 a 3  21 x 2  36 a 5 + 8 a; 6  9 x + 42 a 4  1.
No rational binomial is an exact cube; but, by the rule,
the approximate root may be found.
EVOLUTION.
171
Find, to four terms, the approximate cube roots of the
following :
11. X s + 1.
12. x s —
a°
13. 8
./•
o
CUBE ROOT OF NUMBERS.
244. The method of Art. 243 may he used to extract the
cube roots of numbers.
The cube root of 1000 is 10; of 1000000, is 100; of
1000000000, is 1000; and so on. Hence, the cube root of a
number less than 1000 is less than 10 ; the cube root of a num
ber between 1000000 and 1000 is between 100 and 10 ; the
cube, root of a number between 1000000000 and 1000000 is
betw T een 1000 and 100 ; and so on.
Or, in other words, the integral part of the cube root of a
number of one, two, or three figures, contains one figure; of
a number of four, five, or six figures, contains two figures ;
of a number of seven, eight, or nine figures, contains three
figures ; and so on. Hence,
If a point is placed over every third figure in any integral
a miller, beginning with the units' place, the number of points
will show the number of figures lit the integral part of its eube
root.
245. Let it be required to find the cube root of 405224.
Pointing the number according to
the rule of Art. 244. it appears that
there are two figures in the integral
part of the cube root. Let '/ denote
the figure in the tens' place in the
root, and b that in the units' place.
405224
343000
70 + 4
14700
840
16
15556
62224
62224
Then a must be the greatest mul
tiple of 10 whose cube is less than 405224 ; this we find to
be 70. Subtracting a 3 , that is, the cube of 70, or 343000,
from the given number, we have the remainder 62224. Divid
ing this remainder by 3 a' 2 , or 14700, gives 4, which is the
172 ALGEBRA
value of b. Adding to the trial divisor 3 a b, which is 840,
and b' 2 , which is 16, completes the divisor, 15556. Multiplying
the result by 4, and subtracting the product, 02224, there is
no remainder. Therefore we conclude that 70 + 4, or 74, is
the required cube root.
The work is usually arranged thus :
405224
343
74
14700
840
16
15556
62224
62224
RULE.
Separate the given member into periods, by pointing every
third figure, beginning ivith the units' place.
Find the greatest cube in the lefthand period, and place its
root on the right / subtract the cube of this root from the left
hand period, and to the remainder bring down the next period
for a dividend.
Divide this dividend, omitting the last two figures, by three
times the square of the root already found, and annex the quo
tient to the root.
Add together the trial divisor, with two zeros annexed;
three times the product of the last root figure by the rest of the
root, ivith one zero annexed ; and the square of the last root
figure.
Multiply the divisor, as it now stands, by the figure of the
root last obtained, and subtract the product from the dividend.
If there are more periods to be brought down, continue the
operation in the same manner as before, regarding the root
already obtained as one term.
The observations made after the rule for the extraction of
the square root (Art. 241) are equally applicable to the extrac
tion of the cube root.
EVOLUTION.
173
EXAMPLES.
1 Extract the cube root of 8.144865728.
8.144865728
8
2.012
120000
600
1
120601
144865
120601
12120300
12060
4
24264728
24264728
12132364
Ans. 2.012.
Here it will be observed that, in consequence of tbe in
the root, we annex two additional zeros to the trial divisor,
1200, and bring down to tbe corresponding dividend another
period.
Extract the cube roots of the following :
2. 1860867.
4. 1481.544.
6. 51.478848.
3. .724150792.
29791
681472
7. .000517781627.
Extract the cube roots of the following to the third decimal
place :
8. 3.
10. 212.
12 4
9. 7.
11. 5
8
13 3
13 ' 17
246. JVJien n + 2 figures of a cube root have been obtained
by the ordinary method, n more may be obtained by division
only, supposing 2 n + 2 to be the whole number.
174 ALGEBRA.
Let N represent the number whose cube root is required, a
the part of the root already obtained, x the rest of the root ;
then
$ 2V= a + x,
whence, iV= a 3 + 3 a 2 x + 3 a x 2 + x s ;
therefore, _A r — a 3 = 3 a 2 x + 3 a x 2 + x 3 ,
iV" — a 3 x 2 x 3
» + — +
dec a 6 a
Then A 7 "— a 3 divided by 3 a 2 will give the rest of the cube
root required, or x, increased by  1 77—,; and we shall show
a o a
that the latter is a proper fraction, less than h, so that by
neglecting the remainder arising from the division, we obtain
the part required. For, x by supposition contains n figures,
so that x 2 cannot contain more than 2 n figures. But « con
o
X"
tains 2 n + 2 figures ; and hence  — is less than ^ . And as
n — 5= — X o — , and  — is less than 1,  — . must also be less
o cr a 6 a 6 a 6 a 1
than T T n . Therefore, ( ^ — ^ is a proper fraction, less than \.
a k> a
Remarks similar to those in the last part of Art. 242 apply
here.
ANY ROOT OF POLYNOMIALS.
247. In order to establish a general rule for the extraction
of roots, it will be necessary to notice the formation of the n\\\
power of a polynomial, n being any entire number whatever.
Thus,
(a + by = a n + n a n ~ } b +
Therefore,
y' a n + n a u ~ l b + = a + b.
The first term of the root, a, is the nth root of a n , the first
term of the power; and the .second term of the root, b, may be
EVOLUTION. 175
obtained by dividing the second term of the power by n a n ~\
or by n times the (n — l)th power of the first term of the root.
If the root now found be raised to the nth power, and sub
tracted from the given polynomial, it will be seen that two
terms of the required root have been determined.
It will be observed that the process is essentially that of the
preceding Articles, simplified by dispensing with completed
divisors, and generalized. Hence the •
RULE.
Arrange the terms according to the powers of some letter.
Find the required root of the first term for the first term of
the root, and subtract its power from the given polynomial.
Divide the first term of the remainder by n times the
(n — l)th power of this root, for the second term of the root,
and subtract the nth power of the root now found from the
given polynomial.
If other terms of the root require to be determined, use the
same divisor as before, and proceed in like manner till the nth
poiver of the root becomes equal to the given polynomial.
This rule is, also, applicable to numbers, by taking n figures
in each period.
EXAMPLES.
1. Extract tl* cube root of x 6 + 6 x 5 — 40 x s + 96 x — 64.
x 6 + 6 x 5  40 x 3 + 96 x  64
(* 2 )
2\3 x 6
x 2 + 2x
3 a; 4 1 6 x 5
(x 2 + 2 x) 3 = x 6 +6 x 5 + 12x*+Ssci a
Sx 4 12* 4
(x 2 + 2x4) 3 = x 6 + 6 x 5  40 x 3 + 96 x  64
Ans. x + 2 x — 4.
2. Extract the cube root of ??i 6 — 6 m 5 + 40 m 3 — 90 m — 64.
3. Extract the square root of a i —2a 3 x + 3a 2 x 2 —2ax 3 + x i .
176 ALGEBRA.
4. Extract the fifth root of 32 x 5  80 x* + SO x s — 40 x"
+ 10 a; — 1.
5. Extract the fourth root of x s — 4 a 7 + 10 a; 6 — 16a; 5 + 19 x 4
 16 a 3 + 10 x 1  4 x + 1.
248. When the index of the required root is a multiple of
two or more numbers, we may obtain the root by successive
extractions of the simpler roots.
For, since (Art. 17), ( 7 a) mn = a,
taking the nth. root of both members, we have (Art. 235),
Taking the mth root of both members,
y/ a== y (y a).
Or, the mnth root of a quantity is equal to the mth root of
the nth root of that quantity.
EXAMPLES.
1. Extract the fourth root of 16 x i  96 x 3 y + 216 x 2 y 2
 216 x y s + 81 y\
2. Extract the sixth root of a 12 — 6 a 10 + 15 a 8 — 20 a 6
+ 15 a 4  6 a + 1.
3. Extract the fourth root of m 8 — 8 m 1 ± 12 w 6 + 40 m 5
 74 m*  120 m s + 108 m 2 + 216 wi + 81.
XXII. — THE THEORY OF EXPONENTS.
249. In Art. 17, we defined an exponent as indicating how
many times a quantity was taken as a factor; thus
a m means ay. ay, a to m factors.
Obviously this definition has no meaning unless the expo
nent is a positive integer; and as fractional and negative ox
EXPONENTS. 177
portents have not been previously defined, we may give to
them any definition we please.
250. We found (Arts. 82, 93, and 228) that when m and
n were positive integers,
I. a m Xa n — a m + n .
a m
II. — = a m ~ n .
a n
III. («'")" = a mn .
As it is convenient to have all exponents follow the same
laws, as regards multiplication, division, and involution, we
shall define fractional and negative exponents in such a way
as to make Ride I hold for any values of m and n. We shall
now find what meanings must be assigned to them in con
sequence.
3
251. To find the meaning of a?.
As Rule I is to hold universalh T , it follows that
a a a + a s
a 2 Xa 2 = a 2 2 = a 2 =a 3 .
a
Hence, a 2 is such a quantity as when multiplied by itself
3
produces a 3 . Then, by the definition of root (Art. 17), a 2 must
a
be the square root of a? ; or, a 2 = \J a z .
Again, to find the meaning of a 3 .
i i i. i + i + i a
By Rule I, a 3 X a, 3 X a 3 ■= a 3 3 3 = a 3 = a.
Hence, a* is such a quantity as when taken 3 times as a
factor produces a ; or, a 3 = fya.
252. We will now consider the general case.
p
To find the meaning of a q > p and a being positive integers.
178 ALGEBRA.
p_ p p
By Kule I, a q X a 9 X a 5 X to q factors
P P P P yy
— +—i 1 to? terms — X<2
= a q q « =a q =a p .
p
Hence, a 5 is such a quantity as when taken q times as a
p
factor produces a p . Then a' 1 must be the qi\\ root of a p ;
p
or, a* = ya p .
3 4 5 1
For example, a* = \j a z \ c 2 =\J c 5 ; x 3 ' = y x ; etc.,
and, conversely, y' a 5 = « 4 ; y 7 ic = x 2 ; y" m 5 = m, 3 " ; etc.
EXAMPLES.
253. Express the following with radical signs instead of
fractional exponents :
a 2 x i
a
3.2 c 2 . 5. x*y^. 7. 4a 5 J« 9. 5y T °£ T ^.
2.6^. 4. 3«m^. 6, rn'iA 8. 2c«t/l 10. 3»^c^l
Express the following with fractional exponents instead of
radical signs :
11. yV. 13. sjn. 15. 3 y/ m 5 . 17. ty a 1 ty a?.
12. yV 14. yV. 16. 4 ^a 10 . 18. v^vV
19. 5sjm»%n<. 20. 2 av Vy>.
254. To find the meaning of a 3 .
By Kule I, a~ 3 X a 3 = a = 1, by Art. 94.
Hence, a ~ 3 = —  .
To find the meaning of a~ 2 .
By Rule I, a~* X c$ — a = 1.
Hence, a. 2 = — .
EXPONENTS. 179
255. "We will now consider the general case.
To find the meaning of a~ s , s being integral or fractional.
By Eule I, a~ s X a s = a° = 1.
1
Hence, a s = — .
1 1 2 1
For example, ar 1 = ~\ a 4 — ^?5 a 3 — ~ '■> e * c v
a'
a a*
1 X 2 23
and, conversely, — ^ = « 2 ; — = a; 2 a 3 ; — = 2 a * ; etc.
We observe, in this connection, the following important
principle :
^4 quantity may be changed from the denominator of a frac
tion to the numerator, or from the numerator to the denomi
nator, if the sign of its exponent be changed.
EXAMPLES.
256. Remove all powers from the denominators to the
numerators in the following :
ar 5a; 3 2 a; 1 a; a; 2 , a; 2 x~
a 2 a 3 — 1 a 4 a 5 — b
3.
4.
x 3
3
a 4
+
a; 2 a; 5
7 m
6c 1
3 m
3
7 c*
4m 2 l 3??? 3 + 2?i
i +
5 c 6 2 c" 5
Remove all powers from the numerators to the denominators
in the following :
*o
_ 2a; 3a" 2 a _ a? a: 3 x~ 2 2x 1
180 ALGEBRA.
a 6 3a 4 5a 2 a
7. —^^ — ^ +
8.
a; + 2 5 b 03 7 — b a'
m — x m 3 % 5 2p
1 — x 2 3x 5x 1 7 x
K ™ — 1 7 ^. — 3 *
Express the following with positive exponents:
J_ 2. _3
9. 2x 2 y 2 — 3x~ l y i '— x~ i y r .
10. a 1 5 2 + 2cr 3 i 4 3fth~l
_i _s _i
11. 3a; 3 y 7 — 4 a* ?/ + a y~ •
12. a" 1 6~ 2 c 8 + a~ 2 b~% e 4 + a s j2 c#
257. We obtained the meanings of fractional and negative
exponents on the supposition that Bule I, Art. 250, was to
hold universally. Hence, for any values of m and n,
a m x a n — a m + n
3 2 3_2. J^
For example, a 2 X «~ 5 = a 2 ~ 5 = a~ 3 ; a 4 X « — «* 3 = a ' 5
a i Xa* = a 2 = a 2 ; a 3 X a 5 = a 3 ° = a To I etc.
EXAMPLES.
Multiply together the following :
1. a 8 and a" 1 . 4. c 8 and y'c 2 . 7. rc and n~?.
2. a 2 and a" 2 . 5. a:" 1 and (far 8 . 8. a^ and af*.
3. a x and a~ 5 . 6. m 2 and ^— . 9. 2 c~ * and  3 a tyc s .
EXPONENTS. 181
10. Multiply A ^+2a*3ftHy26 *4a 3_6a £&*.
a% b~ 2 + 2 a^  3 b 2
2 J~2 _ 4 a J _ o a ~% iih
2 a$ & 1 + 4 a* 6"^  6
Aa*b~*S + 12a H 2 .
6  12 a  J b 2 + 18 a %b
2 a$ b 1  20 + 18 a » 6, ^ws.
Note. It should be carefully remembered, in performing examples like
tbe above, that any quantity whose exponent is is equal to 1 (Art. 94).
Multiply together the following :
11. a 2 b~°  2 + a~ 2 b 2 and a 2 b~ 2 + 2 + a~ 2 b 2 .
3. i i ii a , i i
12. a 4 — a? b± + a 4 b' 2  b* and a 4 + b*.
13. a~ 2  2 a" 1 b + b 2  a b 3 and ar s + 2 or 2 b.
14. 3 ar 1 a~ 2 b 1 + or* b~ 2 and 6 a* b 2 + 2 a 2 b + 2a.
15. x~ s f x~ 2 y2x 1 and 2 x 2 y~ l + 2x 8 ! / 2 ±x i y~\
16. x% y~* + 2 + jc" * y* and 2 a;  * y 4 _ 4 A . $ y f + 2 a 2 y 4 .
17. 2 sc^ — 3 a;^ — 4 + aT* and 3 a;'*' + x — 2 x*.
18. 4 a 4 & 1 + « 4  3 «~ 4 6 and 8 a 4 & 1  2 a~±  6 «~ 4 6.
258. To prove that Rule II holds for all values of 'm and n.
By Rule I, a m ~ n X a n = a m ~ n + w = a m .
Inverting the equation, and dividing by a n , we have
— = a"
a n
182 ALGEBRA.
ft 3 cc~ 2
For example, — =:a 3_1 =:a 2 ; — — = a~ 2 ~ 3 = a~ 5 ;
I _ + 2 f a 3 8 + 4 V
a
 = a 4 ' " = a 4 ; —  = a 5 = a 5 ; etc.
EXAMPLES.
Divide the following:
_i _4 l 1
1. a 3 by a 1 . 4. a 2 by a . 7. x by ^— g.
2. a by a 3 . 5. rMiyJc 5 . 8. 5 » by 2 or 1 $ 6.
3. a^ by a* 6. m 2 by tym, 2 . 9. a" 1 6^ by  3 a 6~~*.
10. Divide 2 a^ 6" 1 20 + 18 a  """ 6 by J b~ * + 2 J  3 6*.
2 J' b 1  20 + IS a ^b
2a?b~ 1 +4 : cfib~^—6
2 _l JL I
a :i 6 »+2a* — 3 6'
2 6 2  4 a * — 6 a $ b 2 , ^?*s.
4 a* 6 * — 14 + 18 a ^6
4a^6~^ 8 + 12 »~^ 6^
6 12 a"* 6^+ 18 a~ %b
 6  12 a~^ 62 + 18 <T% b
Note 1. Particular attention must be given to seeing that the dividend
and divisor are arranged in the same order of powers, and that each re
mainder is brought down in the same order. It must be remembered that
a zero exponent is greater than any negative exponent ; and that negative
exponents are the smaller, the greater their absolute value.
Note 2. In dividing the first term of the dividend or remainder by the
first term of the divisor, it will be found convenient to write the quotient
at first in a fractional form; reducing the result by the principles of Art.
258. Thus, in getting the first term of the quotient in Ex. 10, we divide
2 a* b 1 by a? b ' 2 . Then, the result = „ _ L = 2 a 3 3 b + * = 2 b *
(fit *
EXPONENTS. 183
Divide the following :
jl i
11. a — b by a 5 — b b .
12. a 4 + a~ 2 b' 2 + b~ i by a~ 2 + a' 1 b' 1 + b~ 2 .
13. 2x 2 y 2 + 6 + Sx 2 i/ 2 hj2x + 2x 2 y 1 + 4 t x 3 y 2 .
14. 2 x ?s y~ x — 2x~%y + 32 x~ 2 y 3 by 2 + 6 a" § y + 8 as" * ?/.
15. cc~ 3 ?/ 5 — 3 x~ 5 y~ n + x 1 y~ 9 by x~ 2 y~ s + x~ 3 ?/ 4 — a;  4 y 6 .
16. 8 — 10o; 2 2/ J ^+2aj*2/^ > " by 4x~*y% + 2x~ 2 y*—2x~^y 4 .
259. To prove that Rule III holds for all values of m
and n.
We will consider three cases.
Case I. Let m have any value, and n be a positive in
teger.
Then, from the definition of a positive integral exponent,
(a m ) n = a m Xa m Xa m to n factors
— f,m + m + m ton terma __ ~m n
Case II. Let m have any value, and n be a positive frac
v
tion, which we will denote bv  •
P_ q,
Then, (a m ) n = (a m )T= \ (a m ) p , by the definition of Art. 252,
= y 7 ^"^ by Case I, Art. 259,
mp
= aJ, by Art, 252,
= a mx q = a mn .
Case III. Let m have any value, and n be a negative
quantity, integral or fractional, which we will denote by — s.
184 ALGEBRA.
Then, (a m ) n = (a m )~ s — — r s , by the definition of Art. 255,
(a m ) s
= , by Cases I and II, Art. 259,
= a  ms = a m{  s) = a mn .
Thus, we have proved Kule III to hold for all values of m
and n.
For example, (a 2 ) 3 = a 6 ; (a" 1 ) 5 = a" 5 ; (a~ ?s ) * = a"* ;
(J)$ = a; (a*)~* = a ~*J (a 2 )~* = a*; etc.
EXAMPLES.
260. Find the values of the following :
5 '( C " *)">. 10. f ]
1. («•)■ 4. (O  * 7. tf( c 2 ) 2 . 10.  .
2. (« 2 ) 3 . 5. (e*) 2 *. 8. tfm 8 ) * 11. (^
3. (a 3 )*. 6. (,/*)* 9. (^f) 5 . 12. {(^VY 1 
261. To prove that (a £) n = a n b n for any value of n.
In Art. 228 we showed the truth of the theorem when n was
a positive integer.
Case I. Let n be a positive fraction, which we will denote
v E E E
by — . We have then to show that (ab)« =a« b'i.
9.
[(a bfy = (a b) p , by Art. 259.
p p
EXPONENTS. 185
[at &*]* = (c£y (b^) q = a p bP= (ab)", by Art. 228.
Hence, [( a J)fj*=[a? J?] 3 .
Therefore, (a fi) «= a« &«.
Case II. Let w be a negative quantity, which we will de
note by — s. We have then to show that (a b)~ s = a~ 3 b~ 3 .
(a b)° = y—^ = — ;  , by Art. 228 and Case I,
v J (a b) s a s b s J
= ar s b~ s .
262. To find the value of a numerical quantity affected
with a fractional exponent.
1. Find the value of 8*.
From Art. 252, we should have S^^y'S 5 ; and to find the
value in this way, we should raise 8 to the fifth power, and
take the cube root of the result.
A better method, however, is as follows :
8^ = (8*) 5 , by Art. 259,
= ($&)* = 2 s = 32; Ans.
Note. Place the numerator of the fractional exponent as the exponent
of the parenthesis, and 1 divided by the denominator as the exponent of the
cpiantity within.
2. Find the value of 16"*.
 = J__ J_ J_ J_ 1_
~ 16* ~ (IB*)* = ^ 16 ) 5 ~ (± 2 > 5 ~ ±32 :
186
ALGEBRA.
EXAMPLES.
Find the values of the following :
3.27?. 5.1000* 7. (8)1. 9. ('f^f .
36' 2 x 16 *
1 7 s 4^ x 9~ 2
4. 36*. 6. 9"* 8. (27)* 10. ^T 1
81" X 16*
If the numerical quantity is not a perfect power of the de
gree indicated by the denominator of the fractional exponent,
the first method explained in Ex. 1, Art. 262, is the best.
For example, to find the value of 7 2 , we write it \/ 7 s , or
y/ 343 ; and find the square root of 343 to any desired degree
of accuracy.
MISCELLANEOUS EXAMPLES.
263. Extract the square roots of the following :
4 x y c*de 2
5. 9 x* y 2 12x~ s y2 x~ 2 + 4 x~ x y~ x + y~\
6. 4x* + ±x% y~* 15x 2 y~* Sx% y~* + lGx§ y~\
7. z 8 y~% + 6  4 x~% y% + x~ 3 y%  4 x% y~*.
V
Extract the cube roots of the following :
8. ab\ 9. 8a; 4 / 10.
11. $ if  12 y^'x^ 6 y$x 2 i/x 3 .
3m 2 n *
ax 5
EXPONENTS. 187
Reduce the following to their simplest forms :
12. ,^„„^. . 15. a x ~ v+2z a 2x+v ~ 3z a z .
Ji + 2»i+r
13. (x a ) b +(x a ) b . 16. (^'"x^X^ 1 ) *.
ia (a x+v Y ( a? \ x ~ v ,_ r/ _J_\„_»i_2_
Change the following to the form of entire quantities :
18 15 ^* 2 19 X *V 2 20 ^ 2
Reduce the following to their simplest forms :
21. ' ^ 22. ^=^. 23
Factor the following expressions :
24. 9^12^ + 4. 25. a ^3a*88.
26. ar 2 & + 5 a" 1 &* 66.
Factor by the method of Art. 117 :
27. ab. 28. a£&~£. 29. ar'"y — 4«*.
Factor by the method of Art. 119 :
30. a — 6. 31. a + 6. 32. x~ s +8cm^.
188 ALGEBRA.
XXIII. — RADICALS.
264. A Radical is a root of a quantity, indicated by a
radical sign ; as, yfa, \Jx + 1, y m 2 — 2 n + 3.
When the root indicated can be exactly obtained, it is called
a rational quantity ; and when it cannot be exactly obtained,
it is called an irrational or surd quantity.
265. The Degree of a radical is denoted by the index of
© /
the radical sign; thus, \] a is of the second degree; \x + 1
of the third degree.
Similar Radicals are those of the same degree, with the
5/ 6,
same quantity under the radical sign ; as, \ax and 1 y ax.
266. Most problems in radicals depend for their solution
on the following important principle :
For any values of n, a, and b, by Art. 236,
V«X \b = \ab.
REDUCTION OF RADICALS.
TO REDUCE RADICALS OF DIFFERENT DEGREES TO EQUIVALENT
RADICALS OF THE SAME DEGREE.
267. 1. Reduce y 1 2, ^3, and ^ 5 to equivalent radicals of
the same degree.
By Art. 252, ^2 = 2* = 2& =^2" = ^64
{f 3 = 3^ = 3^ = v^ 4 = y'Sl
^5 = 5* = 5& = y' 5 3 = v' 125
RADICALS. 189
RULE.
Express the radicals 10 it h fractional exponents ; reduce these
fractions to a common denominator • express the resulting
fractional exponents with radical signs; and, finally, reduce
the quantities under the radical signs to their simplest forms.
Note. This method affords a means of comparison of the relative mag
nitudes of two or more radicals ; thus, in Example 1, as y/ 125 is evidently
greater than y/Sl, and y'Sl than y/64, hence A'o is greater than y/3, and
^3 than y/2.
EXAMPLES.
Reduce the following to equivalent radicals of the same
degree :
2. s/3, \f4, and ^5. 5. \f2~^ \J3~b, and ^4^
94 ** 6, 4.
3. y 5, y 6, and y 7. 6. y a + b and y a — b.
.3/ , .*/ „ . /— 5 ; ■ 3/
4. SJxy, y x z, and yyz. 7. y'cr — x 2 and sj a z — X s .
8. Which is the greater, ^3 or y'S?
9. Which is the greater, </2 or y/3?
10. Which is the greater, ^4 or $5 ?
TO REDUCE RADICALS TO THEIR SIMPLEST FORMS.
268. A radical is in its simplest form when the quantity
under the radical sign is not a perfect power of the degree
denoted by any factor of the index of the radical, and has no
factor which is a perfect power of the same degree as the
radical.
CASE I.
269. When the quantity under the radical sign is a perfect
power of the degree denoted by some factor of the index of the
radical.
190 ALGEBRA.
a
1. Reduce y 8 to its simplest form.
EXAMPLES.
Reduce the following to their simplest forms :
2. #9. 4. ^27. 6. \/~a^W.
3. ^25 a 2 . 5. $125 a 3 1> 9 . 7. <l 25 "'
CASE II.
270. When the quantity under the radical sign has a
factor which is a perfect power of the same degree as the
radical.
1. Reduce \J 32 to its simplest form.
V 32 = V 16x2 = (Art. 266) ^16x^2 = 4^2, Am.
2. Reduce y/54 a 4 x to its simplest form.
\/5±a i x = \/2Ta s x2ax = % 27~a~ 3 X ^2~a~x = 3 a \/2~a~x~,
Ans.
RULE.
Resolve the quantity under the radical sign into two factors,
one of which is the greatest perfect power of the same degree
us the radical. Extract the required root of this factor, and
prefix the result to the indicated root of the other.
Note. In case the greatest perfect power in the numerical part of the
quantity cannot be readily determined by inspection, it may always be ob
tained by resolving the numerical quantity into its prime factors. Let it
be required, for example, to reduce ^ 1944 to its simplest form. 1944 =
2x2x2x3x3x3x3x3 = 2 8 x3 5 . Hence,
^1944  V / 23lT35 = V2 2 x 3 4 x y/6 = 18 ^6.
RADICALS. 191
EXAMPLES.
Reduce the following to their simplest forms
3. ^50. 6. ^320. 9. 7^63aH 5 c 6 .
4. 3^24. 7. 2^80. 10. %250x 3 fz».
5. si 12. 8. \j98a s b 2 . 11. <Jl8x 3 y i 27 x* y
12. \/ax 2 — 6ax + 9a. 14. \/20 a x 1 + 60 a 2 x + 45 a 3 .
13. SJix^y 2 ) (x + y). 15. ^192 a 4 6 5 + 320 a 3 6 4 .
When the quantity under the radical sign is a fraction, mul
tiply both terms by such a quantity as will make the denomi
nator a perfect power of the same degree as the radical. Then
proceed as before.
/2
16. Reduce t /  to its simplest form.
V / i=\/!=V / (H=Y / i x Y /6 =^ G ' te 
/9
17. Reduce t /  to its simplest form.
Reduce the following to their simplest forms :
18.
19
20
192 ALGEBRA.
TO REDUCE A RATIONAL QUANTITY TO A RADICAL FORM.
271. 1. Eeduce 3 a; 2 to a radical of the third degree.
3 x 2 = \7 (3 x 2 ) 3 = \]21 x\ Ans.
RULE.
Raise the quantity to the power indicated by the given root,
and write it under the corresponding radical sign.
EXAMPLES.
Reduce the following to radicals of the second degree :
a rr a 3x x — 3
2. i a. 3. = . 4. a + 2 x. 5. .
5 x — 2
6. Reduce = to a radical of the fourth degree,
o
TO INTRODUCE THE COEFFICIENT OF A RADICAL UNDER THE
RADICAL SIGN.
272. 1. Introduce the coefficient of 2 a y 3 x 2 under the
radical sign.
2 a V 3a 2 = \/ 8 a 3 X V 3 x 2 = (Art. 266) \f 8 a 3 X 3 x 2 = y 7 24 a 3 x%
Ans.
RULE.
Reduce the coefficient to the form of a radical of the given
degree; multiply together the quantities under the radical
signs, and write the product under the given radical sign.
EXAMPLES.
Introduce the coefficients of the following under the radical
signs :
2. 3^5 4. 4a 2 \/^U. 6. 5c$Ja.
3. 2^7. 5. 3yTT^. 7. ( x l)J(?±^).
RADICALS. 193
ADDITION AND SUBTRACTION OF RADICALS.
273. 1. Find the sum of y/18, \J21, J , and 12 y/^
18
By Art. 270, . ^18 = 3^2
v/27= 3^3
12
2. Subtract ^48 from ^162.
By Art. 270, ^162 = 3^6
^48 =2^6
^6, ^ws.
RULE.
Reduce each radical to its simplest form. Combine the
similar radicals, and indicate the addition or subtraction of
the dissimilar.
EXAMPLES.
Add together the following radicals :
5
3. y'S, ^18, and y/50. 6. ^20, t/i and J
4. ^12, v/48, and v/ 108. 7. y/y/> and y^
5. ^16,^/54, and v' 128. 8. t/i'tf^**^
2
27
7 2
3
194 ALGEBRA.
Subtract the following :
9. v/ 45 from ^ 245. 10. J ~ horn J
Simplify the following :
16
15
11. \/243 a b 2 + V 75 a* +^3a s  54 a 2 b + 243 a b 2 .
12. 7^27^75y/ + v/12y/ly/l.
13. {^16 + 5^54^250^/^ + ^/1 + ^/^. .
15. V 63 « 2 « S4:abx + 2Sb 2 x — ^7d 2 x + 42abx + 63b 2 x.
MULTIPLICATION OF RADICALS.
274. 1. Multiply sj 2 by ^ 5.
^2x^5 = (Art. 266) ^2 X 5 = ^10, Arts.
2. Multiply y' 2 by y' 3.
Reducing to equivalent radicals of the same degree (Art.
267), we have
y/2 X $3 = $8 X \j 9 = ^72, Am.
KULE.
Reduce the radicals, if necessary, to equivalent ones of the
same degree. Multiply together the quantities under the radi
cal signs, and write the product under the common radical
sign.
RADICALS. 195
EXAMPLES.
Multiply together the following :
3. y/ 12 and y/ 3. 6. V 6 a* and \/ 5 a?.
4. ^2 and y^. 7. V^i V^ 4 , and V^^ 6 ) '
5. ^axand^bx. 8. y/ 2, y/ 5, and t/ ^ •
9. Multiply 2y/cc — 3y/?/ by 4y/a: + y/?/.
2 y/cc — 3 \J y
4y/a;+ y/y
8ic — 12 \jxy
+ 2v/^73 2/
8 « — 10 v/a; y — 3 ?/, ^4ras.
Note. It should be remembered that to multiply a radical of the second
degree by itself is simply to remove the radical sigu ; thus,
y/ x x y/ x = x.
Multiply together the following :
10. s/x 2 and y]x + 3. 11. 3y/z  5 and 7 sjx 1.
12. s/.T + lV/zland y'a; + 1 + y/x1 (Art. 106).
13. yV — 1 — a and V «' 2 — 1 + «•
14. \J x — s] y+ \J z and \j x \ \J y — \/ z.
15. y/2y/3 + y/5 and ^2 + ^3 + ^5.
16. 3y/52y/6 + y/7 and 6 y/5 + 4 y/6 + 2 y/7.
17. 4y/35y/22y/5 and 8 y/3 + 10 y/2  4 y/5.
196 ALGEBRA.
Simplify the following :
Square the following (Arts. 104 and 105) :
20. 2^3^/2. 22. \/la 2 +a.
21. 3^8 + 5^3. 23. ^^b\{a~+~b.
DIVISION OF RADICALS.
275. Since (Art, 266), \faXs/b = \fab, it follows that
\a b itya = y / 6.
RULE.
Reduce the radicals, if necessary, to equivalent ones of the
same degree. Divide the quantities under the radical signs,
and write the quotient under the common radical sign.
EXAMPLES.
8/
1. Divide ^ 15 by y/ 5.
Reducing to equivalent radicals of the same degree, we have
^15^5 = ^225^125 = ^/11 = ^/1, Ans.
Divide the following :
2. v' 108 by v' 18. 6. s/2by$/3.
3. V^O^by \l~2Z. ■ 7, % 2 by ^3.
4. v/54b yv /6. 8. s/ 12 hjsj 2.
5. S/lT^by $3~o~. 9. \/Ta~by$Ta~.
RADICALS. 197
INVOLUTION OF RADICALS.
276. 1. Raise fy 2 to the fourth power.
(^ 2) 4 = (2*) 4 = 2* = f 2 4 = ^ 16, Jn».
2. Raise y/ 3 to the third power.
(ft 3) 3 = (3") 3 = 3^ = 3* = \J 3, Ans.
We observe that in the first example the quantity under the
radical sign is raised to the required power ; while in the sec
ond, the index of the radical is divided by the exponent of
the required power. Hence the following
RULE.
If possible, divide the index of the radical by the exponent
of the required poioer. Otherwise, raise the quantity tinder
the radical sign to the required poioer.
Note. If the radical has a coefficient, it may be involved separately.
The final result should be reduced to its simplest form.
EXAMPLES.
3. Raise y/5 to the third power.
4. Square \J1.
5. Find the fourth power of 4 y 3 x.
6. Find the sixth power of y/ a 2 x.
7. Raise \l a — b to the fourth power.
8. Raise 3 a\bx to the fourth power.
9. Find the value of (\/ x + l)\
10. Find the square of 4 V^ 2 — 3.
198 ALGEBRA.
EVOLUTION OF RADICALS.
277. 1. Extract the square root of y 6 x 2 .
\J( y^) = (y 7 ^ 2 )* = {(6 x*)*} * = (6 a; 2 )* = \/~6x~% Ans.
2. Extract the cube root of \/27 x 3 .
V (V27^Q = (V27V)*= {v/(3 z) 3 P~ = {(3 a>)*}* = ( 3 a ')"
= y 3 £, ^4?zs.
We observe that in the first example the index of the radical
is multiplied by the index of the required root ; while in the
second, the required root is taken of the quantity under the
radical sign. Hence the following
RULE.
If possible, extract the required root of the quantity under
the radical sign. Otherwise, multiply the index of the radical
by the index of the required root.
Note. If the radical has a coefficient, which is not a perfect power of
the same degree as the required root, it should he introduced under the
radical sign hefore applying the rule. Thus,
y(i\lax ) = y(\?l6ax ) = ylGax.
The final result should be reduced to its simplest form.
EXAMPLES.
3. Extract the square root of y/2.
4. Find the cube root of y/ 8.
4
5. Find the cube root of \a + b.
6. Find the square root of \x 2 — 2 x + 1.
7. Extract the fifth root of y/32.
8. Extract the cuhe root of y/27.
9. Find the value of ^(3 y/3).
RADICALS. 199
5/
10. Find the fourth root of \ x* y vi .
11. Find the value of \J(±\J2).
TO REDUCE A FRACTION HAVING AN IRRATIONAL
DENOMINATOR TO AN EQUIVALENT ONE
WHOSE DENOMINATOR IS RATIONAL.
CASE I.
278. When the denominator is a monomial.
2 b
1. Reduce ; — to an equivalent fraction whose denominator
sj a
is rational.
Multiplying both terms by y/ a,
2b _2bsja _2bsja
\l a y/ a\j a a
, Ans.
5
2. Reduce ^^ to an equivalent fraction whose denominator
is rational.
Multiplying both terms by y/ 9,
5_ 5^9 _5v^9_5^9
RULE.
Multiply both terms of the fraction by a radical of the same
degree as the denominator, with such a quantity under the
radical sign as will make the denominator of the resulting
fraction rational.
200 ALGEBEA.
EXAMPLES.
Reduce the following to equivalent fractions with rational
denominators :
3 3
3< ^2"
4 1
$2 a
5 5
5 ' IN
CASE II.
6 r 2c
V '27 a 2
279. When the denominator is a binomial, containing only
radicals of the second degree.
1. Reduce — —  r^ to an equivalent fraction whose denomi
nator is rational.
Multiplying both terms by 3 — y/ 2,
10 10( 3^2) fK ^ iA ^ 3010y/2
3T72 = (3 + V2)(3^2) = ^ Art 106 > 7 ' ^
2. Reduce — — to an equivalent fraction whose de
nominator is rational.
Multiplying both terms by ^ 5 + \J 2,
v/5 + v /2 _ ( v /5 + v/2)(y/5 + v/2) _ 7 + 2 v /10
v/5v/2 _ ( v /5 s /2)( v /5 + v /2)~" 3 ; AnS '
RULE.
Multiply both terms of the fraction by the denominator with
the sign between its terms changed.
EXAMPLES.
Reduce the following to equivalent fractions with rational
denominators :
RADICALS. 201
4 2y/5 + y/2 s/a + x+^ax
3< 3TV2* " s/53^2' • y/^r^y^T
a;
, 4y/3 y/«y^ 19 yVly/^+1
2v/3 s/a+s/x ^atl+^at+l
_ VSV 73 o 2 +V« + 1 1Q ^ + V/^ 2 4
«• ~777 To • "• , • A«5 , •
V/2 + v/3 1Va+l » — VaJ 2 4
y/ft + y/5 1Q « — V/V— 1 14 y/x — 4 y/a — 2
sja\jb' ' a + yjd 1 —! ' 2\lx + 3\jx — 2*
280. If the denominator is a trinomial, containing only
radicals of the second degree, by multiplying both terms of
the fraction by the denominator with one of its signs changed,
we shall obtain a fraction which can be reduced to an equiva
lent fraction with a rational denominator by the method of
Case II. Thus, to reduce the fraction
y/2v/3v/7
^2 + ^3 + ^'
Multiplying both terms byy/2 + y/3 — y/7,
v /2 v /3v/7 _ (v/2v/3v/7)(v/2 + v /3v/T) _62 v 14
l/2 + y/3 + y/7~(\/2 + s/3 + \/7)(s/2+sJ3s/7) 2y/62
_ 3y/14
Multiplying both terms by y/ 6 + 1, we Lave
(3 V /14)( v /6 + l) _ 3v/14 + 3y/6 v /84
(v / 6 _l )(v /6 + l)  5
If the denominator is a binomial, containing radicals of any
degrees whatever, it is possible to reduce the fraction to an
equivalent form with a rational denominator ; but the process
is more complicated than the preceding and rarely necessary.
202 ALGEBEA.
281. To find the approximate value of a fraction whose
denominator is irrational, reduce it to an equivalent fraction
whose denominator is rational.
1. Find the value of ^ j^ to three decimal places.
2 J^ = (Art. 279) 2 ±^ 2 = 2 ±^ = 1.707, An*.
It will be seen that the value of the fraction is obtained in
this way more easily than by dividing 1 by 2 — \J 2, or its
value .586.
EXAMPLES.
Find the values to three decimal places of the following :
2 2 3 3 4  5 V /3 V /2
IMAGINARY QUANTITIES.
282. An Imaginary Quantity is an indicated even root of
a negative quantity ; as, y — 4, \/ — a 2 .
In contradistinction, all other quantities, rational or irra
tional, are called real quantities.
283. All imaginary quantities may be expressed in one
common form, which is, a real quantity multiplied by y — 1.
For example,
y/3^ _ yV X (_i) _ (Art. 266) y' a" X ^ :Z 1 = a y^l ;
also, y/ 2 = ^2 X(l) = s/2\[^l.
Hence, we may regard ^—1 as a universal factor of every
imaginary quantity, and use it in our investigations as the
only symbol of such a quantity.
KADICALS. 203
284. Imaginary quantities may be added, subtracted, and
divided the same as other radicals ; but with regard to multi
plication, the usual rule requires some modification.
285. By Art. 17, V— 1 means such an expression as when
multiplied by itself produces — 1 ;
or, (v/^l7 = l;
also, (y/I^^v/^lTxV^T^lV 3 !;
and, (V^1) 4 = (V^1) 2 X(V^ 3 1) 2 =(1)X(1) = 1.
By continuing the multiplication, we should find
(Vl) 8 = l; etc.
Or, in general, where n is any positive integer,
(v/=ir + W=T; (V=i) 4B+2 =i; (V=T) 4n+8 =V^;
(V=1) 4 " +4 = L
MULTIPLICATION OF IMAGINARY QUANTITIES.
286. 1. Multiply ^^a^hy \f^¥.
V^^X V 17 ^^ (Art. 283) a <f^l X ft y'^1 = « ft (V 11 *) 2
= — aft, ^4hs.
2. Multiply V 2 by y^3.
V^2 X y /r 3=v/2x\/3x(v /:: i) 2 =^ ) il
??s.
3. Multiply together ^— 4, V— 9, V~ 16 > and V" 25.
V^ X V 3 ^ X S/^IG x V /Ir 25 = 2x3x4x5x (v^)*
= 120(\/~l) 4 = 120, Arts.
204 ALGEBRA.
RULE.
Reduce all the imaginary quantities to the form of a real
quantity multiplied by ^— 1. Multiply toy ether the real
quantities, and multiply the result by the required power of
EXAMPLES.
Multiply the following :
4. 4^3 and 2\J2. 7. 1 + V 1 and 1  \J 1.
5. _ 3 yL_ a and 4 V b. 8. \J a% V~ b% and y— c 2 .
6. 4 + V 7 and 82^7. 9. a + \/— b and a — \J—b.
10. 2 V 3 — 3 V 2 and 4 V 3 + 6 V~ 2.
11. Divide V « by V~ *•
We should obtain the same result by using the rule of Art.
275 ; hence, that rule applies to the division of all radicals,
whether real or imaginary.
Divide the following :
12. V 6 by V 2.
13. ^^12 by y'S.
Simplify the following :
16. 1 + S ^_1 . (Art. 279).
1Vl 2V2
14.
V
5hj^
1.
15.
tf
 54 by y 7 
17. A + t
2
18. ^£0 + "^ . (Art. 154).
a — \'—b a +^b
19. Expand (2 V^3) a . 20. Expand (2 + 3 Y^2) 8 .
KADICALS. 205
QUADRATIC SURDS.
287. A Quadratic Surd is the indicated square root of an
imperfect square ; as, \J 3, ^ x + 1.
288. A Binomial Surd is a binomial in which one or both
of the terms are irrational.
289. The square root of a rational quantity cannot be
equal to a rational quantity plus a quadratic surd.
If possible, let y' a = b + y/ c
Squaring the equation, a = b 2 + 2 b \j c + c
or, 2 b \J c = a — b 2 — c
. a — b 2 — c
^ C = ^b—
that is, a surd equal to a rational quantity, which is impossible.
290. If tin' sum of a rational quantity and a quadratic
surd be equal to the sum of another rational quantity and
another quadratic surd, the two rational quantities will be
equal, also the two quadratic surds.
That is, if a + \J b — c + \J d
then a = c and \Jb = \J d
For, if a is not equal to c, suppose a = c f x
then c\x\\Jb = c + \Jd
or, x + s/ b = y/ d
which is impossible by Art. 289. Hence, a must equal c, and
consequently y/ b must equal y/ d.
291. To prove that ifsja + ^b — sj x + \J y, then V ' a — \j b
= s /x — sfy.
Squaring the equation \/ a + sjb = \J x + \J y,
we have tc + \Jb — x + 2 \l~r~y + y
Whence, by Art. 290, a = x + y and >Jb = 2 \ xy.
206 ALGEBRA.
Subtracting these two equations, we have
a~sJb — x — 2sJxy\y
Extracting the square root, ^ a — ^b = ^ x — ^ y.
292. To extract the square root of a binomial surd whose
first term is rational.
For example, to extract the square root of a + \J b.
Assume \J a + \fb = \/ x + sj y (1)
then (Art. 291), ^a—sjb = ^ x — \Jy (2)
Multiplying (1) by (2), \ja"b = xy (3)
Squaring (1), a + \/b — x + 2^xy + y
Whence (Art. 290), a = x + y. (4)
Adding (3) and (4), a + \J~aT^b = 2x, or x= l + ^f~ b 
Subtracting (3) from (4), a — SJ a 2 — b=2y, or ?/= ^ .
Substituting these values of x and y in (1) and (2),
^i^=^(l±^EI) + v /(»=^E»). (5)
EXAMPLES.
1. Find the square root of 3 + 2 ^ 2 or 3 + y/ 8.
Here a = 3 and b = 8. Substituting in (5), we have
V 3T^=v/( 3 ±^) + V /('^^)
= v /(^) +v /^H 2 + i ><
RADICALS. 207
2. Find the square root of 6 — \J 20.
Here a = 6 and b = 20. Substituting in (6), we have
293. Examples of this kind may always he solved hy the
following method :
3. Extract the square root of 14 — 4 y/ 6.
y , 14_4 v /6 = V / l^2v /24 = V / 122v / 24 + 2
= (Art. 116)^12^2 = 2^3^2, Arts.
4. Extract the square root of 43 + 15 \J 8.
V43 + 15 si 8 = \M3 + sj 1800 = ^43 + 2 sj 450
= ^25 + 2^450 + 18 = v /25 + V /18 = :5 + 3 v/ 2 > ^ s 
EULE.
Reduce the surd term so that its coefficient may he 2. Sep
arate the rational term into tiro parts whose product shall be
the quantity under the radical sign (see first note on page 48),
writing one part he/ore the surd term and the other part after
it. Extract the square roots of these parts, and connect them
by the sign of the surd term.
The advantage of this method is that it does not require the
memorizing of formulae (5) and (6).
EXAMPLES.
Extract the square roots of the following :
5. 12 + 2^35. 8. 35 + 10 <J 10. 11. 205^12.
6. 242^63. 9. 12 sj 108. 12. 14 + 3^20.
7. 16 + 6^7. 10. 8v/60. 13. 67 7 si 12.
208 ALGEBRA.
Extract the square roots of the following, using formula?
(5) and (6), Art. 292:
14. l12\/2. 15. 7 + 30V2. 16. 353V/16.
17. 2m2\Jmn\ 18. x 2 + a x 2 \/ax 3 .
Extract the fourth roots of the following :
19. 193 + 22 y/ 72. 20. 1712)/ 2. 21. 9756^3.
SOLUTION OF EQUATIONS CONTAINING RADICALS.
CASE I.
294. Wlien there is only one radical term in the equation.
1. Solve the equation v/ar 2 — 5 — x = — 1.
Transposing, ^ x 2 — 5 = x — 1
Squaring, x 2 — 5 = x 2 — 2 cc + 1
Whence, x = 3, Ans.
CASE II.
295. When there are two radical terms in the equation.
2. Solve the equation \] x — \ f x — 3=1.
Transposing, \J x — 1 = ^ x — 3
Squaring, x — 2\Jx + l = x — 3
Transposing and uniting, — 2 y/ x = — 4
or, \jx = 2
Whence, x = 4, Ans.
CASE III.
296. When there are three radical terms in the equation.
3. Solve the equation \J x + 6 + \J x + 13 — v/4a; + 37 = 0.
RADICALS. 209
Transposing, \J x + 6 + \/x + l§ = \/4:X + 3T
Squaring, x + 6 + 2 \/cc 2 + 19.r + T8 + x + 13 = 4 x + 37
Transposing and uniting, 2 y as 2 + 19 a; + 78 = 2 a; + 18
or, ^+19 a; + 78 = a: + 9
Squaring, a; 2 + 19 x + 78 = a;' 2 + 18 a; + 81
Whence, x = 3, ^f «s.
RULE.
297. Transpose the terms of the given equation so that a
radical term may stand alone in one member ; then raise each
member to a power of the same degree as the radical.
If there is still a radical term remaining, repeat the op
eration.
The equation should he simplified as much as possible hefore
performing the involution.
Note. All the examples in tliis chapter reduce to simple equations ;
radical equations, however, may reduce to equations of the second degree,
for the solution of which see Chapter XXIV.
EXAMPLES.
Solve the following equations :
3/
4. V«8 = 3. 6. y'3a; + 4 + 3 = 6. 8. 82^
x
5. V^3 = 2. 7. ^312 = 1. 9. o~\2x = 3.
10. y/4ar192a = l. 14. 6 + ^x = \Jl2 +
11. ^^3^+61=1^. 15. V / '32 + v /ar=16.
12. yx* — 6x 2 + 2 = x. 16. ^x — 3— Va; + 12=— 3.
13. ^ + ^ + 5 = 5. 17. \l2xl+\J2x + 9 = 8.
18. ^3x + 10^3x + 25 = 3.
19. ^x 2 3x + 5\'x 2 5x2 = l.
210 ALGEBEA.
20. \Jx 2 + 4 x + 12 + \jx 2  12 x  20 = 8.
21. v^— ^3 = — .
y x
22. ^3^+^3^+13 = ^ ==.
V o x + 13
23 V 7 ^ — 3 _y/g; — 4
V /tc + T _ "y'ic + 1'
24 y/^_+38_ y/a; + 28
\/# + 6 \J x + 4 '
25. ^1 + ^ + 4=^4^ + 5.
26. yWl + V*2V / 4jc3 = 0.
27. \j2x3 ^8^ + 1 + VlS ^92 = 0.
28. y^  3  V^  14  y/4 a  155 = 0.
29. x \^ (9 + x \/^~~3) = 3.
30. x + l = \f(l + x \/^ r +T6).
31 v5^y/3_v^+_3
V / 2^" v /2~V / ^ + 2'
32. y( a *3a>x + x 2 \/3^~x) = ax.
XXIV. — QUADRATIC EQUATIONS.
298. A Quadratic Equation, or an equation of the second
degree (Art. 164), is one in which the square is the highest
power of the unknown quantity ; as,
ax 2 = b, and x* + 8 x = 20.
299. A Pure Quadratic Equation is one which contains
only the square of the unknown quantity ; as,
ax = b; and a; 2 = 400.
QUADEATIC EQUATIONS. 211
Equations of this kind are sometimes called incomplete equa
tions of the second degree.
300. An Affected Quadratic Equation is one which con
tains both the square and first power of the unknown quan
tity ; as,
x 2 + 8 x = 20 ; and a x 2 + b x — e = b x 2 — a x + d.
Equations of this kind, containing every power of the un
known quantity from the first to the highest given, are some
times called complete equations.
PURE QUADRATIC EQUATIONS.
301. A pure quadratic equation can always he reduced to
the form
x 2 = a,
in which a may represent any quantity, positive or negative,
integral or fractional. Thus, in the equation
20a; 2 , K 9 Jv 41 35x 2
Clearing of fractions, 80 x 2  12 (5 x 2 + 4) = 41  (9  15 x 2 )
or, 80 x 2  60 x 2  48 = 41  9 + 15 x 2
Transposing and uniting terms, 5 x 2 = 80
x 2 = 16
which is in the form x 2 = a.
Equations of this kind have, therefore, sometimes been de
nominated binomial, or those of two terms.
302. An equation of the form
x 2 = a
may he readily solved by taking the square root of each mem
ber. Thus,
x = ± \Ja,
212 ALGEBKA.
where the double sign is used, because the square root of a
quantity may be either positive or negative (Art. 237).
Note. It may seem at first as though we ought to write the douhle sign
before the square root of each member, as follows :
±x = ± y/a.
We do not omit the double sign before the square root of the first member
because it is incorrect, but because we obtain no new results by consid
ering it. The equation ± x = ± y/ a can be written in four different ways,
thus,
x = ^a
— x=tfa
x—  y/«
where the last two forms are equivalent to the first two, and become iden
tical with them on changing all the signs. Hence it is sufficient, in
extracting the square root of both members of an equation, to place the
double sign before one member only.
5x 2
303. 1. Solve the equation 3 x 2 + 7 = —r + 35.
Clearing of fractions, 12 x 2 + 28 = 5 x 2 + 140
Transposing and uniting terms, 7 x 2 = 112
x 2 = 16
Extracting the square root of both members,
x = ± 4, Ans.
RULE.
'Reduce the given equation to the form x 2 = a,and then
extract the square root of both members.
EXAMPLES
Solve the following equations :
2. 4a; 2  7 = 29. 4.
3. 5x 2 + 5 = 3x 2 +5o. 5.
t X 1 —
■5 =
Sx
2
11
5
8
"3
5
4 + x
4
—
X
QUADRATIC EQUATIONS. 213
245
x
= 5x. 7. 13V/3a; 2 +lG = 6.
G
8. x +^x* + 3= ,z—z
y x + 6
9.
_y/3
l_y/l_aj a 1 + y/l
a
.'•
_ cc 2 5 a; 2 _ 7 2 335
10, Y~ + U^2l~ X + ~2l'
11. 2 (x  3) (x + 3) = (x + l) 2  2 x.
12. aa; 2 + 5 = c. 13
x 2 — b x 2 — a
AFFECTED QUADRATIC EQUATIONS.
304. An affected quadratic equation may always be reduced
to the form
x 2 \jpx = q,
where p and q represent any quantities, positive or negative,
integral or fractional. Thus, in the equation
3x — 3 _ 3x — 6
5 x 5 = 2 x H s —
x — J
Clearing of fractions,
10 x (x _ 3) _ (6 x  6) = 4 x (x  3) + (3 x  6) (a;  3)
or, 10 a 2  30 x  6 x + 6 = 4 x 2 — 12 a: + 3 a 2  15 a; + 18
Transposing and uniting terms, 3 a; 2 — 9 x = 12
Dividing by 3, x 2 — 3 x = 4
which is in the form x 2 + p x =■ q.
Equations of this kind have, therefore, sometimes been de
nominated trinomial, or those of three terms.
214 ALGEBRA.
305. Let it be required to solve the equation
x 2 \px = q.
Equations of this kind are solved by adding to both members
such a quantity as will make the first member a perfect square,
and taking the square root of the resulting equation. The pro
cess of adding such a quantity to both sides as will make
the first member a perfect square, is termed Completing the
Square.
In any trinomial square (Arts. 104 and 105), the middle
term is twice the product of the square roots of the extreme
terms ; therefore the square root of the last term must be
equal to half the second term divided hy the square root of
the first. Hence the square root of the quantity which must
be added to x~ + p x to render it a perfect square, is ~ — x,
or — . Adding to both members the square of ^, or ^, we have
2 , _,, , P , p _±q+p
X<+px +  = q + ^ = 
Extracting the square root of both members,
X + P = ± \/±<I+P'
or, x = — 7 ±
p ^iq+jr
2  1  2
Thus, there are two values of x,
p \ 4: q + p' 2 ■ p SJ 4 q + p 2
x = t>+ —o i or
2 ' 2 ' 2 2
We observe from the preceding investigation that the quan
tity to be added to complete the square is found by taking half
the coefficient of cc, and squaring the result.
Hence, for solving affected quadratic equations, we have the
following
QUADKATIC EQUATIONS. 215
RULE.
Reduce the equation to the form x 2 + p x = q.
Complete the square by adding to both members the square
of half the coefficient of x. Extract the square root of both
members, and solve the simple equation thus found.
1. Solve the equation x 2 — 3 x = 4.
Completing the square, by adding to both members the
.3 9
square of , or ,
o 9 . 9 25
x 2 — 3x +  = 4:+  = —
4 4 4
■^ 3 5
Extracting the square root, x — ~ = ± ^
™ 3 5
Transposing, a; =  ± 
Taking the upper sign, x = r+ k = h — 4.
Z ju Li
3 5 2
Taking the lower sign, x= 7> — ^ = — 7 , = — 1.
Ll Li Li
Ans. x = £ or — 1.
We may verify these values as follows :
Putting x = 4 in the given equation, 16 — 12 = 4.
Putting x = l, 1 + 3 = 4.
These results being identical, the values of x are verified.
2. Solve the equation 3 x 2 + 8 x = — 4.
8x4
Dividing through by 3 x 2 + — =— 
o o
216 ALGEBRA.
Completing the square, by adding to both members the square
,4 16
0f 3' 0r y
2 8^ 16 _ 4 16_4
4 2
Extracting the square root, x +  = ± 
o o
m 4 2
1 ransposing, x = —  ± 
rr i • n 4 2
lakmg the upper sign, x = —  +  = — 
Taking the lower sign, x = — ^ — ^ = — ^ = — 2.
2
.4ns. cc = —  or — 2.
o
3. Solve the equation — 3x 2 — 7 x = =.
o
7 x 10
Dividing through by — 3, x 1 + ~ = — —
Completing the square, by adding to both members the square
. 7 49
of 6'° r 36'
X 2
+
7x 49 10 49 __ 9
X + 36" ~~9~ + 36~36
Extracting the
square
root,
7_ 3
X ~t~ „ it A
b b
Transposing,
7 3
X ~ 6 ± 6
Whence,
2 5 A
x = — q or — k ) 4«&
«5 O
A SECOND METHOD OF COMPLETING THE SQUARE.
306. Although any affected quadratic equation may be
solved by the method of Art. 305, since its rule is general,
QUADRATIC EQUATIONS. 217
still it is sometimes rilore convenient to employ a second
method of completing the square, known as the " Hindoo
Method."
An affected quadratic, reduced to three terms, and cleared
of all fractions, may he reduced to the form
a x 2 + b x = c.
Multiplying each term by 4 a, we have
By an operation similar to that of Art. 305, we may show
that b 2 must be added to both members, in order that the first
member may be a perfect square. Thus,
4 a 2 x + 4 a b x + b 2 = b 2 + 4 a c
Extracting the square root, 2 ax \ b = ± \ b 2 \ 4ta c
Transposing, 2 ax = — b ± \ b 2 + 4 a c
tv a i q h±\?b 2 +4ac
Dividing by 2 a, x = ^ .
It will be observed that the quantity necessary to complete
the square, is the square of the coefficient of x in the given
equation. Hence the following
RULE.
Reduce the equation to the form a x 2 + b x = r.
Multiply both members of the equation by four times the
coefficient of x 2 , and add to each the square of the coefficient
of x in the given equation.
Extract the square root of both members, and solve the sim
ple equation thus produced.
Note. The only advantage of this method over the preceding is in
avoiding fractions in completing the square.
4. Solve the equation 2 x 2 — 7 x = — S.
218 ALGEBRA.
Multiplying both members by four times 2, or 8,
16 x 2 5Gx =  24
Adding to each member the square of 7, or 49,
16 x 2 — 56 x + 49 =  24 + 49 = 25
Extracting the square root, 4 x — 7 = ± 5
Transposing, 4cc = 7±5 = 12or2
Dividing by 4, x = 3 or  , ^4«s.
307. This method is usually to be preferred in solving
literal equations.
5. Solve the equation x 2 + (a — 1) x = a.
Multiplying both members by four times 1, or 4,
4 x 2 + 4 (a — 1) x = 4 a
Adding to each member the square of a — 1, or (a — l) 2 ,
4 x 2 + 4 (a  1) x + (a  l) 2 = 4 a + (a  l) 2
= a 2 + 2 a + 1 = (a + l) 2
Extracting the square root,
2 a; + (a — 1) = ± (a + 1)
Transposing, 2 cc = — (a — 1) ± (a + 1)
Taking the upper sign, 2 x = — (a — 1) + (a + 1)
r= — « + l + «+l=2
or,
x = l.
Taking the lower sign,
2 x = (al)(a + l)
= — a + 1 — a — 1 = — 2 a
or,
x — — a.
Ans. x = 1 or — a.
308. In case the coefficient of x in the given equation is
an even number, the rule may be modified as follows :
QUADRATIC EQUATIONS. 219
Multiply both members of the equation by the coefficient of
x 2 , and add to each the square of half the coefficient of x in
the given equation.
6. Solve the equation 7 x 2 + 4 x = 51.
Multiplying both members by 7, 49 x 2 + 28 x = 357
Adding to each member the square of 2, or 4,
49 x 2 + 28 x + 4 = 361
Extracting the square root, 7 x + 2 = ± 19
Transposing, 7 a; — — 2 ± 19 = 17 or — 21
17
Dividing by 7, ic = — or — 3, ^l»s.
SOLUTION OF QUADRATIC EQUATIONS BY A
FORMULA.
309. In Art. 30G, we showed that if a x 2 + b x = c, then
b± \Jb 2 + ±ac ...
X = 2a * (1)
We may use this as a formula for the solution of quadratic
equations as follows :
7. Solve the equation 3 x 2 + 5 x = 42.
Here a = 3, b = 5, c = 42 ; substituting these values in (1),
5±\/25 + 504
X —
6
5±v/529_5±23
8. Solve the equation 110 « 2 — 21 x = — 1.
Here a = 110, 6 = — 21, c = — 1 j substituting in (1),
21 ±1^441 440 21 ±1 1 1 .
X = ~ 22CT  = 220 = I6 Or ll'^
220 ALGEBRA.
Note. Particular attention must be paid to the signs of the coefficients
in substituting.
9. Solve the equation, — x 2 — 6 x = 8.
Here a = — 1, b = — 6, c = 8; substituting in (1),
6 ±^36 32 6±2
x = —jr = — = — 4 or — 2, Ans.
RULE.
Reduce the equation to the form a x 2 + b x = c.
The value of x is then a fraction, ivhose numerator is the
coefficient of x with its sign changed, plus or minus the square
root of the sum of the square of said coefficient, and four times
the product of the second member by the coefficient of x' 2 ; and
whose denominator is twice the coefficient of x 2 .
310. The following equations may he solved hy either of
the preceding methods, preference being given to the one best
adapted to the example considered. Special methods and
devices may also be employed whenever any advantage can
thereby be gained.
EXAMPLES.
Solve the following equations :
10. x° + 2x + 7 = 4:2. 16. 26x + lox 2 = 7.
11. a; 2  9 x 22 = 0. 17.  40 + x = 6 x\
12. x 2 Sx = 15. 18. 17x = 2x 2 6.
13. z 2 +18* = 65. 19. ^+*=_1.
t 7 9 f 2
14. Gx*+7 X 3 = Q. 20. x =^.
£ o o
3x 2 °2
15. 13 z  14 = 3 a;'. 21. ~^ = x .
5 o
QUADRATIC EQUATIONS. 221
22. ^_"_£ = a 24. (as3)(2as + l)=4.
6 Jo
23. ^^ = ^. 25. (*+5)O5)(llx+l)=0.
26. 4x(18xl) = (10xl) 2 .
27. (3a:5) 2 (> + 2) 2 = 5.
28. (asl) 2 (3as + 8) 2 =(2as + 5) 2 .
29 2 as_ 5 _21 ^_ 3 4
29< x + 2~2 d7 " 5as 7 7*
x x — 1_3 _ ft cc + 1 a? + 3 _8
30  ^TI— "F""2' ^+2 x + 4 _ 3
.t 5 — as 15 on 3 a? 2 1 — 8 x x
31. 3 = j . OV.
5 — x . x 4' ' x — 7 10 5
5 3z + l 1 . n 2asl 3x 1
. =  . 40. 1 —
x a; 2 4" x 3 a; — 12
cc
33. = — — . 41. \/20 + xx 2 = 2(x5).
3x+4 4x+l T
34. _ x __— —  = 0. 42. as+V5as + 10 = 8.
3a; + 4 7 as 4
._ . 353as ,. AQ as*a* + 7 n
35. 6 as H = 44. 43. = a; + —
a; ar + 3 a; — 1 o
14 — r 7 3 22
36. 4 a =t,? = 14 44. ^ ^=^
a; + 1 x 2 — 4 a; + 2 o
45. ^— + = —   +
x* — 1 3 3 (as — 1) a; + 1
222 ALGEBRA.
._ a + 3 x — 3 2 a; — 3
46. s+ s = r 
a + 2 x — 2 x — 1
,„ 35 + 2 a — 2 2a + 16
47. T H == =,
a — 1 a + 1 a + 5
12 + 5 a 2 + a 1
12 — 5a a 1 — 5a
49.
\J X
+ 1
— ya
1
a
\Jx
X ■+
4/9
+ 1
■ V2
+ v^
1
"2*
50.
(5x4
 V^
3) =
= 9.
K1
a 2 + 36 a
a
52. a ex 2 — b ex + a d x — b d.
53. a 2  2 a a + a 2  & 2 = 0.
2 a (a — a) a
54
3a2a 4
1 111
55.  =  +  + .
a + b + x a b a
56. (3 a  2) (a + 5)  (a  6) (5 a  16) =301.
57. (2 a + 3) (3 a + 4) = (8 + a) (2 a + 9).
58. (2 a  5) 2  (2 a  1) 2 = 8 a  5 a 2  5.
59. x + b x + c x = (a + c) (a — b).
Sax 6 a 2 + a b  2 b 2 b 2 x
60. abx 2 +
c 2
61. (3 a 2 + b 2 ) (a 2  a + 1) = (3 6 2 + a 2 ) (a 2 + a + 1).
PROBLEMS. 223
XXV. — PROBLEMS
LEADING TO PURE OR AFFECTED QUADRATIC EQUATIONS
CONTAINING BUT ONE UNKNOWN QUANTITY.
311. 1. I bought a lot of flour for $ 175 ; and the number
of dollars per barrel was to the number of barrels, as 4 to 7.
How many barrels were purchased, and what was the price of
each ?
Let x = the number of dollars per barrel,
7 x
then  = the number of barrels.
4
7 x 2
By the conditions, —r = 175
Whence, x = ± 10.
Only the positive value is applicable, as the negative value
does not answer to the conditions of the problem.
That is, x = 10, the number of dollars per barrel,
7 x
and  = 171, the number of barrels.
4
2. There is a certain number, whose square increased by 30,
is equal to 11 times the number itself. Required the number.
Let x — the number.
By the conditions, x 2 + 30 = 11 x
Solving this equation, x = 5 or 6.
That is, the number is either 5 or G, for each of these values
satisfies the conditions of the problem.
3. I bought a watch, winch I sold for $ 56, and thereby
gained as much per cent as the watch cost me. Required
the amount paid for it.
Let x = the amount paid, in dollars.
Then x = the gain per cent,
x X"
and — — X x = — —  = the whole gain in dollars.
224 ALGEBRA.
x 2
By the conditions, ^r— = 56 — x
J 100
Solving this equation, x = 40 or — 140.
Only the positive value of x is here admissible, as the nega
tive result does not answer to the conditions of the problem.
The cost, therefore, was $ 40.
Note. When two answers are found to a problem, they should be ex
amined to see whether they answer to the conditions of the problem or not.
Only those which answer to the conditions should be retained.
PROBLEMS.
4. I have three square houselots, of equal size. If I were
to add 193 square rods to their contents, they would be equiv
alent to a square lot whose sides would each measure 25 rods.
Required the length of each side of the three lots.
5. There are two square fields, the larger of which contains
25,600 square rods more than the other, and the ratio of their
sides is as 5 to 3. Required the contents of each.
6. Find two numbers whose sum shall be 15, and the sum
of their squares 117.
7. A person cut and piled two ranges of wood, whose united
contents were 26 cords, for 356 dimes ; and the labor on each
of them cost as many dimes per cord as there were cords in its
range. Required the number of cords in each range.
8. A grazier bought a certain number of oxen for $ 240, and
having lost 3, he sold the remainder at $8 a head more than
they cost him, and gained $59. How many did he buy ?
9. The plate of a rectangular lookingglass is 18 inches by
12, and is to be framed with a frame all parts of which are of
equal width, and whose area is to be equal to that of the glass.
Required the width of the frame.
10. A merchant sold a quantity of flour for §.'59, and gained
as much per cent as the flour cost him. What was the cost of
the flour ?
PROBLEMS. 225
11. There are two numbers whose difference is 9, and
whose sum multiplied by the greater is 266. What are the
numbers ?
12. A and B gained by trade $ 1800. A's money was in
the firm 12 months, and he received for his principal and gain
$2600. B's money, which was $3000, was in the firm 16
months. What money did A put into the firm ?
13. A merchant bought a quantity of flour for $ 72, and
found that if he had bought 6 barrels more for the same
money, he would have paid $> 1 less for each barrel. How
many barrel's did he buy, and what was the price of each ?
14. A square courtyard has a gravelwalk around it. The
side of the court wants 2 yards of being 6 times the breadth
of the gravelwalk, and the number of square yards in the walk
exceeds the number of yards in the perimeter of the court by
164. Required the area of the court.
15. My gross income is $ 1000. After deducting a percent
age for income tax, and then a percentage, less by one than
that of the income tax, from the remainder, the income is
reduced to 8 912. Required the rate per cent at which the
income tax is charged.
16. The sum of the squares of two consecutive numbers is
113. What are the numbers ?
17. Find three consecutive numbers such that twice the
product of the first and third is equal to the square of the
second, increased by 62.
18. I have a rectangular field of corn which consists of 6250
hills ; and the number of hills in the length exceeds the num
ber in the breadth by 75. How many hills are there in the
length and breadth ?
19. A certain company agreed to build a vessel for $ 6300 ;
but, two of their number having died, those that survived had
each to advance $ 200 more than they otherwise would have
done. Of how many persons did the company at first consist ?
226 ALGEBRA.
20. A detachment from an army was marching in regular
column, with C men more in depth than in front; hut Avhen
the enemy came in sight, the front was increased by 870 men,
and the whole was thus drawn up in 4 lines. Required the
number of men.
21. A has two square gardens, and the side of the one ex
ceeds that of the other by 4 rods, while the contents of both
are 208 square rods. How many square rods does the larger
garden contain more than the smaller ?
22. A certain farm is a rectangle, whose length is twice its
breadth; but should it be enlarged 20 rods in length and 24
rods in breadth, its contents would be doubled. Of how many
acres does the farm consist ?
23. A square courtyard has a rectangular gravelwalk
around it. The side of the court wants one yard of being six
times the breadth of the gravelwalk, and the number of square
yards in the walk exceeds the number of yards in the perim
eter of the court by 340. What is the area of the court and
width of the walk ?
24. A merchant bought 54 bushels of wheat, and a certain
quantity of barley. For the former he gave half as many
dimes per bushel as there were bushels of barley, and for the
latter 4 dimes per bushel less. He sold the mixture at $ 1 per
bushel, and lost $ 57.60 by his bargain. Required the quan
tity of barley, and its price per bushel.
25. A lady wishes to purchase a carpet for each of her square
parlors ; the side of one of them is 1 yard longer than the
other, and it will require 85 sqxiare yards for both rooms.
What will it cost the lady to carpet each of the rooms with
carpeting 40 inches wide, at 81.75 per yard ?
26. A man has two square lots of unequal dimensions, con
taining together 15,025 square feet. If the lots were contigu
ous to each other, it would require 530 feet of fence to embrace
them in a single enclosure of six sides. Required the area of
each lot.
QUADRATIC EQUATIONS. 907
27. A certain number consists of two digits, the lefthand
digit being twice the righthand ; and if the digits are inverted,
the product of the number thus formed, increased by 11, and
the original number, is 4956. Find the number.
28. A man travelled 108 miles. If he had gone 3 miles
more an hour, he would have performed the journey in 6 hours
less time. How many miles an hour did he go ?
29. A cistern can be filled by two pipes running together in
2 hours 55 minutes. The larger pipe by itself will fill it
sooner than the smaller by 2 hours. What time will each pipe
separately take to fill it ?
30. A set out from C towards D, and travelled 3 miles an
hour. After he had gone 28 miles, B set out from D towards
C, and went every hour ^ of the entire distance ; and after
he had travelled as many hours as he went miles in an hour,
he met A. Required the distance from C to D.
31. A courier proceeds from P to Q in 14 hours ; a second
courier starts at the same time from a place 10 miles behind P,
and arrives at Q at the same time as the first courier. The
second courier finds that he takes half an hour less than the
first to accomplish 20 miles. Find the distance from P to Q.
XXVI. — EQUATIONS IN THE QUADRATIC
FORM.
312. An equation is in the quadratic form when it is ex
pressed in three terms, two of which contain the unknown
quantity ; and of these two, one has an exponent twice as
great as the other. As,
x 6 — 6 x 3 = 16,
x s + x* = 72,
(x  1) 2 + 3 {x  1) = 18, etc.
228 ALGEBRA.
313. The rules already given for the solution of quadratics
will apply to equations having the same form. For, in the
equation
a x 2n + b x n = c,
let x n = y ; then x 2n = if. Substituting,
ay 2 +by = c
Whence, hy Art. 309, we have
or, x
•
— b± \Jb 2 + kac
U —
2a
/*n —
— b±\Jb 2 + 4tac
2 a
from which equation x may he found by extracting the wth
root of both members.
314. 1. Solve the equation sc 4 — 5 x 2 = — 4.
The equation may be solved as in Art. 313, by representing
x by y. A better method, however, is the following :
Completing the square, x* — 5.r 2 + r = — 4= + ~r — t
Extracting the square root, • x 2 —  = ± 
Transposing, a; 2 = 7 r± = 4orl
Whence, x = ± 2 or ±1, Ans.
2. Solve the equation x 6 — 6 x s = 16.
Completing the square, x 6 — 6 x a + 9 = 16 + 9 = 25
Extracting the square root, x a — 3 = ± 5
Transposing, cc 3 = 3±5 = 8or — 2
Whence, x = 2 or — \/2, Ans.
QUADRATIC EQUATIONS. 229
Here, although the equation is of the sixth degree, we find
hut two roots. The equation in reality has six roots, hut this
method fails to give more than two. It will he shown here
after how to obtain the other four.
3. Solve the equation x + 4 \Jx = 21.
Writing the radical with a fractional exponent,
x + 4x^ = 21
which is in the quadratic form.
Completing the square, cc + 4^:r + 4 = 21 + 4 = 25
Extracting the square root, \jx + 2 = ± 5
Transposing, y/x = — 2 ± 5 = 3 or — 7
Whence, squaring, x = 9 or 49, Arts.
7 1
4. Solve the equation 3 x 2 + x^ = 3104 x*.
Dividing hy x^, 3x% + x% = 3104
which is in the quadratic form.
Multiplying hy four times 3, or 12,
36 x% + 12 x% = 37248
Completing the square, 36 x% + 12 x% + 1 = 37249
Extracting the square root, 6 x® + 1 = ± 193
Transposing, 6 x% = — 1 ± 193 = 192 or  194
5 97
Dividing hy 6, x* = 32 or — y
\ /97\i
Extracting the fifth root, x b = 2 or — ( jrJ
Raising both members to the sixth power,
x =64 or (4)°> Ana.
230 ALGEBRA.
EXAMPLES.
Solve the following equations :
k t
5. .x 4 + 4cc 2 = 117. 11. 3 a;* ^ = 592.
6. a" 4  9 or" 2 + 20 = 0. 12. a 3 a: 2 =56.
7. a 10 + 31 a 5 10 = 22. 13. x2s/x = 0.
8. 81 x + 4 = 82. 14. x% + x* = 756.
x
9. *• + !??? 14 = 60. IS. ^+2 = i=V»
a; 2 4 + ya: ya:
3^*
2
10. a; 6 + 20 a; 3 10 = 59. 16. — — = ^.
x — 5 20
17. Solve the equation (x  5) 3  3 (x  5) 2 = 40.
a 9 9 169
Completing the square, (a; — 5) 3 — 3 (x — 5) 2 +  = 40 +  = — —
a 3 ^3
Extracting the square root, (x — 5)* — ■= — ± =
3 3 13
Transposing, (a; — 5) ^ =  ± — = 8 or — 5
Squaring both members, (x — 5) 3 = 64 or 25
Extracting the cube root, x — 5 = 4 or \J 25
Whence, x = 9 or 5 + \J 25,
QUADRATIC EQUATIONS. 231
Solve the following equations :
18. (x 2 oa : ) 2 8(x 2 5x)=84:.
19. (2 x  1) 2  2 (2 x  1) = 15.
20. (3 x 2) 2  ll(3a; 2 2) + 10 = 0.
21. (;r J 5) 2 + 29 3 5)=:96.
22. Solve the equation x* + 10 x 3 + 17 x 2 — 40 aj — 84 = 0.
We may write the equation in the form
x* + 10 x 3 + 2ox 2 — 8x 2 A0x = 84
or, (;z 2 +5a;) 2 8(cr + 5a)=84
Completing the square, (x 2 +ox) 2 —8(x 2 +5x) + 16 = 100
Extracting the square root, (x 2 + 5 x) — 4 = ± 10
Transposing, (a; 2 + 5 x) = 4 ± 10 = 14 or — 6
Taking the first value, we have x 2 + 5 x = 14
Whence (Art. 309), x = ~ 5 ±^ 5 + 56 = =M? = 2 or  7.
Taking the second value, we have x 2 + 5 x = — 6
w, _5±v/2524 5±1
Vv hence, a; = ^ =  = — J or — 3.
Ans. x = 2, — 7, — 2, or — 3.
Note. In solving equations of this form, our object is to form a perfect
trinomial square with the ,r* and x 3 terms, and a portion of the x 2 term.
By Art. 305, we may effect this by separating the x 2 term into two parts, one
of which shall be the square of the quotient obtained by dividing the x 3
term by twice the square root of the x* term.
232 ALGEBRA.
Solve the following equations :
23. x* 12 x 3 + 3±x 2 + 12cc = 35.
24. x i + 2x 3 2o x  26 x + 120 = 0.
25. x 4  6 cc 3  29 x 2 + 114 x = 80.
26. a 4 + 14 x 3 + 47 x  14 a;  48 = 0.
27. Solve the equation 2 ar + \'2 x 2 + 1 = 11.
We may write the equation, (2 x 2 + 1) + y 2 x' 2 + 1 = 12
49
Completing the square, (2 x 2 + 1) + V 2 x 2 + 1 +  = j
1 7
Extracting the square root, V^^+l + T^i^
_ 1 7
Transposing, V 2 ^ + * = ~ 2 ± 2 =r3 ° r_4
Squaring, 2 a; 2 + 1 = 9 or 16
Transposing, 2 x 2 = 8 or 15
15
Dividing by 2, a; 2 = 4 or ~^
/15
Whence, x = ± 2 or ± t/ — , ^4?is.
Note. In solving equations of this form, add such quantities to hoth
members, that the expression without the radical in the first member may
be the same as that within, or some multiple of it.
Solve the following equations :
28. 2 x 2 + 3 x  5^2^+3 x + 9 = 3.
29. x  6 x + 5 \?x 2  6 x + 20 = 46.
30. 4 z 2 + 6 \/4 ar + 12 x  2 =  3 (1 + 4 as).
31. x 2  10 a:  2 yV 2  10 ^ + 18 + 15 = 0.
32. 3.r 2 +15a:2v/ar"+5a: + l = 2.
QUADRATIC EQUATIONS. 233
XXVII.— SIMULTANEOUS EQUATIONS
INVOLVING QUADRATICS.
315. The most general form of an equation of the second
degree containing two unknown quantities, is
ax 2 +bxy+cy 2 + dx + ey + f= 0,
where the coefficients a, b, c, etc. represent any quantities,
positive or negative, integral or fractional.
316. Two equations of the second degree containing two
unknown quantities will generally produce, hy elimination, an
equation of the fourth degree containing one unknown quantity.
Thus, if the equations are
x 2 + y = a
x + y 2 = b
From the first, hy transposition, y = a — x 2 ; substituting in
the second,
x + (a — x 2 ) 2 = b
or, x* — 2 a x 2 + x + a 2 — b =
an equation of the fourth degree. The rules for quadratics
are, therefore, not sufficient to solve all simultaneous equations
of the second degree.
In several cases, however, their solution may he effected hy
means of the ordinary rules.
CASE I.
317. WJien each equation is of the form a x 2 + b y 2 = c.
1. Solve the equations,
3x 2 + Aij 2 = 7Q
3 y 2  11 x 2 ^ 4
234 ALGEBRA.
Multiplying the first equation by 3, and the second by 4,
9 x 2 + 12 if = 228
12 fUx 2 = 16
Subtracting, 53 x 2 = 212
x 2 = ±, x = ±2.
Substituting these values in either given equation,
When x = 2, y = ± 4.
When a? = — 2, y = ±4.
^4«s. # = 2, ?/ = ± 4 ; or, x = — 2, y = ± 4.
EXAMPLES.
Solve the following equations :
2. 2 a; 2 + y 2 = 9 ; 5 x 2 + 6 y 2 = 26.
3. 4 a; 2  3 f =  11 ; 11 x 2 + 5 y 2 = 301.
4. 9 x 2 + 24 v/ 2 = 7 ; 72 a; 2  180 ?f —  37.
5. 20 x a  16 y 2 = 179 ; 5 x 2  336 ?/ 2 = 24.
CASE II.
318. When one equation is of the first degree.
1. Solve the equations,
x 2 + y 2 = 13
x + y — 1
From the second, by transposition, y = 1 — x (1)
Substituting in the first, x 2 + 1 — 2 x + x 2 = 13
or, cc 2 — x =6
AVI / A * o no \ • 1 ± S/T+2l 1 ± 5 _
\\ hence (Art. 309), cc = L  = —  — = 3 or — 2.
Substituting these values in (1),
When a = 3, y = l — 3 — — 2.
W T hen x = 2, y = l + 2 = 3.
Ans. x = 3, y = — 2 ; or, x = — 2, y = 3.
QUADKATIC EQUATIONS. 235
In solving examines under Case II, we find an expression
for the value of one of the unknown quantities in terms of the
other from the simple equation, which we substitute for that
quantity in the other equation, thus producing a quadratic
containing only one unknown quantity, by means of which
the values of the unknown quantities are readily obtained.
Note. Although some examples, in which one equation is of the first
degree (Ex. 1 for instance), may be solved by the methods of the next case,
yet the method of Case II will be found in general the simplest.
EXAMPLES.
Solve the following equations :
2. x\r y= — 1; xy = — 56.
. 3. x + y = 3 ; x 2 + y 2 = 29.
4. x s — y 3 = — 37 ; x — y = — 1.
5. x — y = s; xy = 20.
6. 10cc + y = 3ccy;3/ — cc = 2.
7. x — y = 5; xy = — 6.
8. x 3 + y 3 = 9 ; x + y = 3.
9. 3x 2 2icy = 15; 2« + 3y = 12.
10. x  y = 3 ; x 2 + y" = 117.
11. x + y — 11; xy = 18.
12. x — y = 6; x 2 + y* = 90.
13. x 3 + y 3 = 152 ; x + y = 2.
14. x 2 + 3 x y — xf = 23 ; x + 2 y = 7.
15. x 3 y 3 = 9S; xy = 2.
16. x + y = ±; x 2 + y 2 = 58.
CASE III.
319. When the given equations are symmetrical icith
respect to x and y.
236 ALGEBRA.
1. Solve the equations,
x" + y 2 = 68
x y = 16
Multiplying the second by 2, 2 x y = 32
Adding this to the first equation, x 2 + 2 x y + y' 2 = 100 (1)
Subtracting it from the first equation,
x 2 2xy + y 2 =36 (2)
Extracting the square root of (1), x + y = ± 10 (3)
Extracting the square root of (2), x — y — ± 6 (4)
Equations (3) and (4) furnish four pairs of simple equations,
x + y = 10 x + y — 10 x + y = — 10 x + y = — 10
x—y—6 x—y=—Q x—y=6 x—y=—6
2x = 16
2x = 4,
2x = A
2x = 16
x = 8.
x = 2.
x = — 2.
x = — 8.
y = 2.
y = 8.
y = S.
y = 2.
Ans. x = 8, y = 2; x — 2,y = S;
x = — 2, y = — 8 ; or, x = — 8, y = — 2.
2. Solve the equations,
a;3 + ^ = 133
x 2 — x y + y 2 = 19
Dividing the first equation by the second,
x + y = 7 (1)
Squaring (1), x 2 + 2 x y + y 2 = 49 (2)
Subtracting the second given equation from (2),
3 x y = 30 ; or, 4 x y = 40 (3)
Subtracting (3) from (2), x 2  2 x y + y 2 = 9
Whence, x — y=±3 (4)
QUADRATIC EQUATIONS. 237
Adding (1) and (4), 2 x = 10 or 4
Whence, x = 5 or 2.
Substituting these values in (1),
When x = 5, y = 2
cc = 2, y = 5.
Ans. x — 5, y = 2 ; or, a; = 2, ?/ = 5.
The example might have been solved by substituting the
value of y derived from (1) in either of the given equations,
as in Case II.
The student will notice the difference between Examples 1
and 2 as regards the arrangement of the last portion of the
work.
3. Solve the equations,
x + y = 20
Multiplying the first equation by 2, 2 x 1 + 2 y 2 = 416 (1)
Squaring the second equation, x 2 + 2 x y + y 2 = 400 (2)
Subtracting (2) from (1), x 2  2 x y + y 2 = 16
Whence, x — y = ± 4 (3)
Adding the second given equation and (3),
2 x = 24 or 16
Whence, x = 12 or 8.
Substituting these values in (3),
When » = 12, y = 8
x = 8, y = 12.
J??5. aj .= 12, y = 8 ; or, a; = 8, y = 12.
This example is solved more readily by the method of Case
II; we solve it by Case III merely to show how equations
may be solved symmetrically, when one is of the first degree.
238 ALGEBRA.
EXAMPLES.
Solve the following equations :
4. x' 2 + if = 25 xy = 12.
5. x* + y 2 = S5; xy = 42.
6. x 3 + y 3 = 19; x 2 xy + y 2 = 19.
7. x 3  y 3 =  65 ; x 2 + x y + y 2 = 13.
8. o; + 2/ = l;a;y = — 6.
9. cc 2 + y 2 = 65 ; a; — y = 11.
10. a? 9 +y 2 = 61; x + y = ll.
11. a; 8 — y 8 = 117; a; — y = 3.
Note. Exs. 8, 9, 10, and 11 are to be solved like Ex. 3, and not by
the method of Case II. In solving Ex. 11, begin by dividing the first
equation by the second.
CASE IV.
320. When the equations are of the second degree, and
homogeneous.
Note. Some examples, in which both equations are of the second de
gree and homogeneous, are solved more easily by the methods of Cases I
and III, than by that of Case IV. The method of Case IV is to be used
only when the example can be solved in no other way.
1. Solve the equations,
x 2 — xy = 35
xy + y 2 = 18
Letting y = vx, we have
35
x 2 — v x 2 = 35, or x 2 (1 — v) = 35 ; whence, x 2 = —   — (1)
18
v x 2 + v 2 x 2 = 18, or x 2 (v + v 2 ) = 18 ; whence, x 2 =  '■ — r
v + v
QUADRATIC EQUATIONS. 239
Equating the values of x 2 , z = =
10 1 — vv+v 2
Clearing of fractions, 35 v + 35 v 2 =18 — 18 v
Transposing and uniting, 35 v 2 + 53 v = 18
Whence (Art. 309),
 53 ± V '809 + 2520  53 ± 73 2 9
V = 70 = 70 =7° r 5
2
If v =  , substituting in (1), x 2 = 49, or x = ± 7
Substituting in the equation y = v x,
o
When x = 7, y=xT=2
x = 7, !/ = ~x7 = 2.
If v = — ■=, substituting in (1), x 2 =  7 r , ov x = ± —j^
£> 2 y 2
Substituting in the equation y = vx,
«n 5 9 5 9
men x =V2' y= ~5 x 72 = ~72
5 9 5 9
^2'^ 5' N V 2 V 2
.4ns. cc = 7, ?/ = 2; £ = — 7, y = — 2;
5_ 9 j5_ _9_
X ~^/2 ,y ~ S/2 ] ° l,X ~ \J2 ,y ~s/2'
Note. In using the equation y~v x, to calculate the value of y when
x has been found, care should be taken to use that value of v which n/i*
used in getting the particular value of x.
EXAMPLES.
Solve the following equations :
2. x 2 + xy + ±y 2 = 6; 3x 2 +8y 2 = U.
3. 6x 2 5xy + 2y 2 = 12; 3 x 2 + 2 x y 3 y 2 = 3.
240 ALGEBRA.
4. x 2 + x y = 12 ; x y — y 2 = 2.
5. 2 y 2  4 x y + 3 x 2 = 17 ; y 2 x 2 = 16.
6. x 2 + x y — y 2 = 1 ; x 2 — x y + 2 y 2 = 8.
7. 2 x 2 — 2 a; y — ?/ 2 = 3; x 2 + 3a;y + 2/ 2 = 11.
321. We append a few miscellaneous examples, for the
solution of which no general rules can be given. Various arti
fices are used ; familiarity with which can only be obtained by
experience.
1. Solve the equations,
x z — y s = 19
x 2 y — x y 2 = 6
Multiplying the second by 3, 3 x 2 y — 3 x y 2 = 18 (1)
Subtracting (1) from the first given equation,
x z — 3 x 2 y + 3 x y 2 — y 3 = 1
Extracting the cube root, x — y = 1 (2)
Transposing, x = 1 + y (3)
Dividing the second given equation by (2), x y = 6 (4)
Substituting from (3) in (4), y (1 + y) = 6
or, y 2 + y = 6
m l±V/T+24 1±5
Vv hence, ?/ — k — i5 — = ^ or — 3.
Substituting in (3),
When y = 2, x = 3
y = — 3 } x = — 2.
Ans. x = 3, y = 2 ; or, a; = — 2, ?/ = — 3.
QUADRATIC EQUATIONS. 241
2. Solve the equations,
V *
x + y = 12
Let x — u + v, and y = u — v.
Then x + y = 2 u ; whence, 2u — 12, or u = 6.
From the first given equation, x z + y s = 18 x y
Substituting x = 6 + v, and y = 6 — v, we have
(6 + vf + (6  v) 3 = 18 (6 + v) (6  1>)
Reducing, 432 + 36 v 2 = 648  18 v 2
Whence, 54v 2 = 216
v 2 = 4, v = ± 2
Then a5 = 6+v = 6±2 = 8or4.
Substituting these values in the second given equation,
When x = 8, y = 4
a; = 4, ?/ = 8, ^4ws.
3. Solve the equations,
x 2 + y" + x + y = 18
x ?/ = 6
Adding twice the second equation to the first,
x 2 + 2 x y + y 2 + x + y — 30
or, (x + y) 2 +(;x + y)=30
N  1 ± Vl + 120 1±11 r r
Whence, (x + y) = ^ = „ = 5 or — 6.
Taking the first value, x + y = 5 (1)
and the second given equation, xy = 6 (2)
From (1), y=5—x ; substituting in (2), x 2 — 5 x = — 6
5±\/2524 5±1
Whence, x = —^ = — „ — = 3 or 2.
Substituting in (1),
When, x = 3, y — 2
cc = 2, y = S.
242 ALGEBRA.
Taking the second value, x + y = — 6 (3)
and the second given equation, x y = 6 (4)
From (3), y — — C — x ; substituting in (4),
x 2 + 6 x = — 6
_6±y/3624 6+2^3 /0
Whence, a; = ^_ = =r_VL_ — _ 3 ± ^ 3.
Substituting in (3),
When a= — 3 + v/ 3 ; t/ = — S—^3.
x = 3sJ3,y = — 3 + sJ3.
Ans. x = 3, y = 2 ; cc = 2, ?/ = 3 ;
a; = — 3+\/3, y=— 3— ^3; or, x=— 3— y/3, ?/=— 3 + y/3
4. Solve the equations,
x* + 7/ = 97 (1)
* +y =~1 (2)
Raising (2) to the fourth power,
x i + 4 x s y + 6 x 2 y 2 + 4 x y 3 + y 4 = 1 (3)
Subtracting (1) from (3), 4 a? 3 y + 6 ar ?/ 2 + 4 a; ?/ 3 = — 96
or, 3 a; 2 .?/ 2 + 2 a; ?/ (x 2 + ?/ 2 ) =  48 (4)
But from (2), x 2 + y 2 = l2xy
Substituting in (4), 3 x 2 y 2 + 2 x y (1  2 x y) =  48
or, a: 2 y 2 — 2 cc y = 48
Whence, x y = = — — — = — 6 or 8.
Taking the first value, x y = — 6
From (2), y = — 1 — x ; substituting, x 2 + x = 6
Whence, x = j — = —  = 2 or — 3.
Substituting in (2),
When x = 2, y = — 3.
x = — 3, y = 2.
Taking the second value, x y = 8
QUADRATIC EQUATIONS. 243
From (2), y = — 1 — x ; substituting, x 2 + x = — 8
l±y/T^32 l±y/3i
\\ hence, x = = = „ .
Substituting in (2),
 1 + V /=7 31  1  V^3l
When x = ^ > V~ <j '
 1  y/^31  1 + \/^31
*= — 2 — >y= —2 —
Ans. x = 2,y = ~3; xS,y = 2; x=  1+ ^~ 31 ^
_1_ S /Z3i _i_y/Z3i _i +v /I73i
y = y — i ov > x = 2~ >y= 2 '
EXAMPLES.
Solve the following equations :
5. x + y = 9; \/x+\/y = 3..
6. x + \J~x~y + y = 19 ; x 2 + x y + f = 133.
7. a; 2 y + ar y 2 = 30 ; x* if + x~y i = 468.
8. x 2 + y 2 — 'x  y = 18 ; x y + x + y = 19.
9. x 2 + 3x + y = 73  2 x y ; y 2 + 3 y + x = 44.
. « n „ 5 x 1/ xy
10. as» + I f=_£j x  y = f.
♦ J 4
„, a; 4v/# 33 „
11.  + 7— = r ; » — y=&
y \jy 4
12  2 + 8 = li 5 + S= 4
13. x 2 y + x y 2 = 30 ; a 3 + y s = 35.
14. cc+V / *'?/=3;?/+V«2/ = — 2
15. x 2 y + y 2 x = 6;  +  = .
16. r 4 + ?/ 4 = 17; x — y = 3.
17. z 5  y 5 =  211 ; cc — y = — 1.
18. ar + y 2 = 7 + as y ; x s + y s = 6 x y — 1.
19. 2x 2 7x?/2?/' 2 =:5; 3x yx 2 + 6 y 2 = U.
244 ALGEBRA.
20. A^ + Jl^S  +  = 2.
y + 3 x + J 2 2 o
21. z + 2 = 7; 2y3* = 5; a; 2 + y 1  z~ = 11.
22. xz = y 2 ; (x + y)(z—x—y) = 3; (x + y + z) (z— x— y) = 7.
XXVIII. — PROBLEMS
LEADING TO SIMULTANEOUS EQUATIONS INVOLVING
QUADRATICS.
322. 1. What two quantities are those, the sum of whose
squares is 130, and the difference of whose squares is 32 ?
Let x = one number,
and y = the other.
By the conditions, x 2 + y 2 = 130
x 2 y 2 = 32
Solving these equations, as in Case I, Art. 317,
x = 9, y = ± 7 ; #
or, x = — 9, y = ± 7.
This indicates four answers to the problem :
9 and 7,
9 and — 7,
— 9 and 7,
— 9 and — 7.
Any one of these pairs of values will satisfy the conditions
of the problem.
2. A says to B, " The sum of our money is $ 18." B re
plies, "If twice the number of your dollars were multiplied by
PROBLEMS. 245
mine, the product would be $ 154." How many dollars had
each ?
Let x = A's dollars,
and y = B's.
By the conditions, x + y = 18
2 x y = 154
Solving these equations, as in Case II, Art. 318,
x = li y = U;
or, x = 11, y = 7.
That is, either Alias $7, and B $11, or A has $11, and
B $7.
3. The price of two coats and one vest is $ 38. And the
price of a coat less that of a vest, is to $ 23, as $ 7 is to the
sum of the prices of a coat and vest. What is the price of a
coat, and what of a vest ?
Let x = the price of a coat in dollars,
and y = the price of a vest.
By the conditions, 2 x + y = 38
and x — y : 23 = 7 : x + y
or (Art. 181), a 2 ?/ 2 = 161
Solving these equations, as in Case II, Art. 318,
x = 15, y = 8 ;
107 100
x = — ,y =  — .
Only the first answer is admissible, as a negative value of
either unknown quantity does not answer to the conditions
of the problem. Hence, the price of a coat is $15, and of a
vest, $ 8.
Note. The note after Ex. 3, Art. 311, applies with equal force to the
problems in this chapter.
246 ALGEBRA.
PROBLEMS.
4. The difference of two quantities is 5, and the sum of
their squares is 193. What are the quantities ?
5. There are two quantities whose product is 77, and the
difference of whose squares is to the square of their difference
as 9 to 2. Required the quantities.
6. A and B have each a field, in the shape of an exact
square, and it requires 200 rods of fence to enclose hoth. The
contents of these fields are 1300 square rods. What is the
value of each at $ 2.25 per square rod ?
7. Two gentlemen, A and B, were speaking of their ages.
A said that the product of their ages was 750. B replied, that
if his age were increased 7 years, and A's were diminished 2
years, their product would be 851. Required their ages.
8. A certain garden is a rectangle, and contains 15,000
square yards, exclusive of a walk, 7 yards wide, which sur
rounds it, and contains 3696 square yards. Required the
length and breadth of the garden.
9. What k two numbers are those whose difference multi
plied by the less produces 42, and by their sum, 133 ?
10. A and B lay out money on speculation. The amount
of A's stock and gain is $ 27, and he gains as much per cent
on his stock as B lays out. B's gain is $ 32 ; and it appears
that A gains twice as much per cent as B. Required the capi
tal of each.
11. I bought sugar at such a rate, that the price of a pound
was to the number of pounds as 4 to 5. If the cost of the
whole had been 45 cents more, the number of pounds would
have been to the price of a pound as 4 to 5. How many pounds
were bought, and what was the price per pound ?
12. A and B engage in speculation. A disposes of his share
for $ 11, and gains as many per cent as B invested dollars.
PROBLEMS. 247
B's gain was $ 36, and the gain upon A's investment was 4
times as many per cent as upon B's. How much did each
invest ?
13. A man bought 10 ducks and 12 turkeys for $ 22.50. He
bought 4 more ducks for $ 6, than turkeys for $ 5. What was
the price of each ?
14. A man purchased a farm in the form of a rectangle,
whose length was 4 times its breadth. It cost £ as many dol
lars per acre as the field was rods in length, and the number
of dollars paid for the farm was 4 times the number of rods
round it. Required the price of the farm, and its length and
breadth.
15. I have two cubic blocks of marble, whose united lengths
are 20 inches, and contents 2240 cubic inches. Required the
surface of each.
16. A's and B's shares in a speculation altogether amount
to $ 500. They sell out at par, A at the end of 2 years, B of
8, and each receives in capital and profits $ 297. How much
did each embark ?
17. A person has $ 1300, which he divides into two portions,
and loans at different rates of interest, so that the two por
tions produce equal returns. Jf the first portion had been
loaned at the second rate of interest, it would have produced
$ 36 ; and if the second portion had been loaned at the first
rate of interest, it would have produced $49. Required the
rates of interest.
18. Two men, A and B, bought a farm of 104 acres, for
which they paid $320 each. On dividing the land, A says to
B, "If you will let me have my portion in the situation which
I shall choose, you shall have so much more land than I, that
mine shall cost $ 3 per acre more than yours." B accepted the
proposal. How much land did each have, and what was the
price of each per acre ?
248 ALGEBRA.
19. A and B start at the same time from two distant towns.
At the end of 7 daj's, A is nearer to the halfway house than
B is, by 5 miles more than A's day's journey. At the end of
10 days they have passed the halfway house, and are distant
from each other 100 miles. Now it will take B 3 days longer
to perform the whole journey than it will A. Required the
distance of the towns, and the rate of walking of A and B.
20. Divide the number 4 into two such parts that the prod
uct of their squares shall be 9.
21. The forewheel of a carriage makes 15 revolutions more
than the hindwheel in going 180 yards ; but if the circumfer
ence of each wheel were increased by 3 feet, the forewheel
would only make 9 revolutions more than the hindwheel in
going the same distance. Find the circumference of each
wheel.
22. A ladder, whose foot rests in a given position, just
reaches a window on one side of a street, and when turned
about its foot, just reaches a window on the other side. If
the two positions of the ladder are at right angles to each
other, and the heights of the windows are 36 and 27 feet re
spectively, find the width of the street and the length of the
ladder.
23. A and B engaged to reap a field for 90 shillings. A
could reap it in 9 days, and they promised to complete it in 5
days. They found, however, that they were obliged to call in
C, an inferior workman, to assist them the last two days, in
consequence of which B received 3.?. 9d. less than he other
wise would have done. In what time could B and C each reap
the field?
24. Cloth, being wetted, shrinks J in its length and ,\ ; in its
width. If the surface of a piece of cloth is diminished by 5^
square yards, and the length of the four sides by 4} yards,
what was the length and width of the doth originally?
THEORY OF QUADRATIC EQUATIONS. 249
XXIX. — THEORY OF QUADRATIC EQUA
TIONS.
323. A quadratic equation cannot have more than two
roots.
We have seen (Art. 304) that every complete quadratic
equation can be reduced to the form
x 2 + p x = q.
Suppose, if possible, that a quadratic equation can have three
roots, and that r 1; r 2 , and r 3 are the roots of the equation
x° + p x — q. Then (Art. 166),
r 1 2 +pr 1 = q (1)
r 2 2 +pr 2 — q (2)
r 3 2 +pr 3 — q (3)
Subtracting (2) from (1), (?y — r.F) + p (r x — r 2 ) =
Dividing through by r y — r 2 , which by supposition is not
zero, as the roots are not equal,
r \ + r 2 +p =
Similarly, by subtracting (3) from (1), we have
n + n + p — o
Hence, r x + r 2 + p = r x + r s f p
or, r 2 = r 3 .
That is, two of the roots are identical. Therefore, a quad
ratic equation cannot have more than two roots.
DISCUSSION OF THE GENERAL EQUATION.
324. By Art. 305, the roots of the equation x 2 + p x = q
are
p\ \/p 2 + ±q 1 —p — \Jp 2 +4:q
250 ALGEBRA.
1. Suppose q positive.
Since p 2 is essentially positive (Art. 227), the quantity
under the radical sign is positive and greater than ,p° ; so
that the value of the radical is greater than P Hence, one
root is positive, and the other negatn r e.
If p is positive, the negative root is numerically the larger;
if p is zero, the roots are numerically equal ; and if p is nega
tive, the positive root is numerically the larger.
2. Suppose q equal to zero.
The quantity under the radical sign is now equal to p 2 ; so
that the value of the radical is p. Hence, one of the roots is
equal to 0. The other root is positive when p is negative, and
negative when p is positive.
3. Suppose q negative, and 4 q < p 2 .
The quantity under the radical sign is now positive and less
than p> 2 ; so that the value of the radical is less than p.
If p is positive, hoth roots are negative ; and if p> is nega
tive, both roots are positive.
4. Suppose q negative, and 4 q =p 2 .
The quantity under the radical sign is now equal to zero ;
so that the two roots are equal ; being positive if p is negative,
and negative if p is positive.
5. Suppose q negative, and 4 q > p 2 .
The quantity under the radical sign is now negative ; hence,
by Art. 282, both roots are imaginary.
325. All these cases may be readily verified by examples.
Thus, in the equation x 2 — 3;r = 70, as p is negative and q
positive, we should expect to find one root positive and the
other negative, and the positive root numerically the larger
And this is actually the rase, for on solving the equation, wfc
find x = 10 or  7.
THEORY OF QUADRATIC EQUATIONS. 251
326. From the quadratic equation x 2 +px = q, denoting
the roots by r x and r. 2 , we have
p+S/Y + lq _ 1m pSjp' + lq
2
n = — ^ , and r 2
Adding these together, we have
2p
r x + r 2 = — ^ = —p.
Multiplying them together, we have
n r 2 = £=M±±1± (Art. 106) =  *± =  q .
That is, if a quadratic equation be reduced to the form
x 2 + p x = q, the algebraic sum of the roots is equal to the co
efficient of the second term, with its sign changed • and the
product of the roots is equal to the second member, with its
sign changed.
327. The equation a x 2 + b x + c = 0, by transposing c, and
dividing each term by a, becomes
x 2 \
bx c
a a
Denoting the roots of the equation by x l and x 2 , we have, by
the previous article,
b _c
Xi j Xo — , and x± x.y — — •
a a
328. A Quadratic Expression is a trinomial expression of
the form a x 2 + b x + c. The principles of the preceding
article enable us to resolve any quadratic expression into two
binomial factors.
The expression a x' 2 + b x + c may be written
f bx c
a I x \ 1 —
V a a
252 ALGEBRA.
b c
By the previous article,  = —(% + x 2 ), and  =x 1 x 2 , where
x x and x 2 are the roots of the equation ax 2j rbx + c = Q;
which, we ohserve, may he obtained by placing the given
expression equal to 0. Hence,
ax 2 + bx + c = a [x 2 — (x x + x 2 ) x + x x x 2 ~\.
The expression in the bracket may be written
/V»** rtn /yt ry* /y> I /y» sy
which, by Case II, Chap. VIII, is equal to (x — x^ (x — x 2 ).
Therefore, a x 2 + b x + c = a (x — x x ) (x — x 2 ).
1. Factor 6 x 2 + 11 x + 3.
Placing the expression equal to 0, and solving the equation
thus formed, we find
_  11 + ^121  72 _  11 ± 7 _ 3 1
X ~~ ~V1~ ~ _ ~~12  ~2' 0r ~3
Then, a = 6, x 1 = — ^, x 2 = — 5.
Therefore, 6 x 2 + 11 x + 3 = 6 (x + ) (» + 5)
= (2 a; + 3) (3 x + 1), Ans.
2. Factor 4 + 13 x  12 x 2 .
Placing the expression equal to 0, and solving the equation
formed, we have
_  13 ± y/ 169 + 192 _  13 ± 19 _ 4 1
X ~ 24 24 ~3' ° r 4
4 1
Then, a = — 12,x 1 = 7 r, x 2 = — .
THEORY OF QUADRATIC EQUATIONS. 253
Therefore, 4 + 13 x  12 x 2 =  12 he  ^J (a; + ^J
= 3{xl)4(x + \)
— (4 — 3 x) (4 a; + 1), ^4?is.
Note. It should be remembered, in using the formula a (xx{) (x  a'2 N ,
that a represents the coefficient of x 2 in the given expression J hence, in
Example 2, we made a=  12.
EXAMPLES.
Factor the following expressions :
3. x 1 + 73 x + 780. 9. 8z 2 + 18z5.
4. x 2  11 x + 18. 10. 4 z 2  15 * + 9.
5. x 2 4cc60. 11§ 2x 2 +x6.
6. a 8 + 10 a; 39. 12. 9x 2 12a + l.
7. 2 a; 2 7 a; 15. 13. l8xx 2 .
8. 21 a 2 + 58 a; + 21. 14. 49 x 2 + 14 a  19.
329. The principles of Art. 328 furnish a method of form
ing a quadratic equation which shall have any required roots.
For, the equation a x 2 + b x + c = 0, if its roots be denoted
by x x and x 2 , may be written, by Art. 328,
a
(x — Xy) (x — x 2 ) = 0, or (x — £Cj) (x — a 2 ) = 9.
Hence, to form an equation whose roots shall be x v and x.,,
we subtract each of the two roots from x, and place the product
of the resulting binomials equal to zero.
7
1. Required the equation whose roots are 4 and — r •
By the rule, (x — 4) (x + )=
254
or,
ALGEBRA.
x 2 
9x
7 =
=
4x 2 
9x
28 =
= o,
Ans.
Clearing of fractions,
EXAMPLES.
Form the equations whose roots are
17
2. 1 and — 2. 5. 7 and — 6 \ . 8. — ^ and 0.
8 4
3. 4 and 5. 6. —  and  . 9. 1 + sj 5 and 1 — ^5.
o i
4. 3 and —■=. 7. — 2 J and — 3^. 10. m + \/ w and m — \Jn.
o
330. By Art. 328, the equation a x 2 + b x + c = may be
written (x — a^) (x — x 2 ) = 0, if acj and x 2 are its roots; we
observe that the roots may be obtained by placing the factors
of the first member separately equal to zero, and solving the
simple equations thus formed.
This principle is often useful in solving equations.
1. Solve the equation (2x — 3) (3 x + 5) = 0.
3
Placing the first factor equal to zero, 2x — 3 = 0, or a; =  .
A
5
Placing the second factor equal to zero, ox+ 5=0, orx= —  .
A 3 5
Ans. x =  or — ^ .
A o
2. Solve the equation x' 2 + 5 x = 0.
The equation may be written x (x + 5) =
Placing the first factor equal to zero, x = 0.
Placing the second factor equal to zero, x + 5 = 0, or x = — 5.
Ans. x = or — 5.
THEORY OF QUADRATIC EQUATIONS. 255
EXAMPLES.
Solve the following equations :
3. (as) (as2) = 0. 9. 2as 8 18as = 0.
4. (as+.5)(asl) = 0. 10. (2as + 5)(3asl) = 0.
5. (as?) (as + ?)=0. 11. (aas + b) (cxd) =0.
6. (as + 8)(as + i)=0. 12. (x 2 4) (as 2 9) = 0.
7. 2as 2 13as = 0. 13. (3 a; + 1) (4 x*  25) = 0.
8. 3 as 3 + 12 a: 2 = 0. 14. (as 2 a)(as 2 aas£)=0.
15. as (2 as + 5) (3 x 7) (4 x + 1) = 0.
16. (x 2 5x + 6)(x 2 + 7x+ 12) (2 x* + 9x5) = 0.
331. Many expressions may be factored by the artifice of
completing the square, used in connection with the method of
Case IV, Chapter VIII.
1. Factor x i + a\
x* + a 4 = x* + 2 x 2 a 2 +a*2 x 2 a 2
= (x 2 + a 2 ) 2  (a x ^/ 2) 2
= (Art. 117) (x 2 + a x ^ 2 + a 2 ) (x 2 — a x \J 2 + a 2 ), Ans.
2. Factor x 2 — ax + a 2 .
as 2 — ax + a 2 = x 2 + 2ax + a 2 — 3ax.
= (x + a) 2 (^3axY
= (x + V^3 a x + a) (x — \J3 a x + a), Ans.
256 ALGEBRA.
EXAMPLES.
Factor the following expressions :
3. x 2 + l. 5. a 2 3ab + b*. 7. x 2 xl.
4. x 2 +x+l. 6. x i lx 2 y 2 ^y\ 8. m i + m 2 n 2 +?i\
332. We have seen (Art. 330) that any equation whose
first member can be factored, and whose second member is
zero, may be solved by placing the factors separately equal to
zero and solving the equations thus formed. This method of
solution is frequently the only one which will give all the roots
of the equation.
1. Solve the equation x 3 = 1.
The equation may be written x 3 — 1 = 0, or (Art. 119),
(x  1) (x 2 + x + 1) = 0.
Placing the first factor equal to zero,
x — 1 = 0, or x = 1.
Placing the second factor equal to zero,
x 2 + x + 1 = 0, or x 2 + x = — 1
Whence (Art, 309),   = ^ ± ^
x =
2 2
Hence, x = 1 or ^1 , Ans.
EXAMPLES.
Solve the following equations :
2. x i = l. 4. x 4 +« 4 = 0. 6. x 6 = l.
3. x 3 = l. 5. X 4 X * + 1 = 0. 7. rr 4  — +1 = 0.
These examples afford an illustration of the statement made
in Art. 167 that the degree of an equation indicates the num
ber of its roots.
DISCUSSION OF PROBLEMS. 257
XXX. — DISCUSSION OF PROBLEMS
LEADING TO QUADRATIC EQUATIONS.
333. In the discussion of problems leading to quadratic
equations, we find involved the same general principles which
have been established in connection with simple equations
(Arts. 205212), but with certain peculiarities.
These peculiarities will be now considered. They arise from
two facts :
1. That every quadratic equation has two roots • and
2. That these roots are sometimes imaginary.
334. In the solution of problems involving quadratics, it
has been observed that the positive root of the equation is
usually the true answer ; and that, when both roots are posi
tive, there may be two answers, either of which conforms to
the given conditions.
The reason why results are sometimes obtained which do
not apply to the problem under consideration, and are there
fore not admissible, is that the algebraic mode of expression
is more general than ordinary language ; and thus the equa
tion which conforms properly to the conditions of the problem
will also apply to other conditions.
1. Find a number such that twice its square added to three
times the number may be 65.
Let x = the number.
Then 2 x 2 + 3 x = 65 (1)
13
Whence, x = 5 or — .
The positive value alone gives a solution to the problem in
the sense in which it is proposed.
To interpret the negative value, we observe that if we
change x to — x, in equation (1), the term 3 x, only, changes
258 ALGEBRA.
its sign, giving as a result the equation 2 x' 2 — 3 x = 65. Solv
13
ing this equation, we shall find x = — or — 5, which values
only differ from the others in their signs. We therefore may
13
consider the negative solution, — , taken independently of
Li
its sign, the proper answer to the analogous problem (Art.
205):
" Find a number such that twice its square diminished by
three times the number may be Go."
2. A farmer bought some sheep for $ 72, and found that if
he had bought 6 more for the same money, he would have paid
$ 1 less for each. How many sheep did he buy ?
Let x = the number of sheep bought.
72
Then — = the price paid for one,
x
72
and = the price paid, if 6 more.
JO \~ O
72 72
By the conditions, — =  + 1
J ' x x + 6
Whence, x = 18 or — 24.
Here the negative result is not admissible as a solution of
the problem in its present form; the number of sheep, there
fore, was 18.
If, in the given problem, "6 more " be changed to "6 fewer"
and "$1 less" to "$1 more," 24 will be the true answer.
Hence, we infer that
A negative result, obtained as one of the answers to a prob
lem, is sometimes the answer to another analogous problem,
formed by attributing to the unknown quantity a quality
directly opposite to that which has been attributed to it.
DISCUSSION OF PROBLEMS. 259
INTERPRETATION OF IMAGINARY RESULTS.
335. It lias been shown (Art. 324) under what circum
stances a quadratic equation will be in form to produce imagi
nary roots. It is now proposed to interpret such results.
Let it be required to divide 10 into two such parts that their
product shall be 26.
Let x = one of the parts.
Then 10 — x = the other.
By the conditions, x (10 — x) = 26
Whence, x = 5 ± y/— 1.
Thus, we obtain an imaginary result. We therefore con
clude that the problem cannot be solved numerically ; in fact,
if we call one of the parts 5 + y, the other must be 5 — y, and
their product will be 25 — y' 2 . which, so long as y is numerical,
is less than 25. But we are required to find two numbers
whose sum is 10 and product 26 ; there are, then, no such
numbers.
Had it been required to find two expressions, whose sum is
10 and product 25. the answer 5 + \/ — 1 and 5 — \j— 1
would have satisfied the conditions.
The given problem, however, expresses conditions incom
patible with each other, and, consequently, is impossible.
Hence,
Imaginary results indicate that the problem is impossible.
PROBLEM OF THE LIGHTS.
336. The principles of interpretation will be further illus
trated in the discussion of the following general problem.
Find upon the line which joins two lights, A and B, the
point which is equally illuminated by them : admitting that
the intensity of a light, at a given distance, is equal to its
2G0 ALGEBRA.
intensity at the distance 1, divided by the square of the given
distance.
C" A C B C
— I 1 J 1 1 —
Assume A as the origin of distances, and regard all dis
tances estimated to the right as positive.
Let a denote the intensity of the light A, at the distance 1 ;
b the intensity of the light B, at the distance 1 ; and c the
distance A B, between the two lights.
Suppose C the point of equal illumination, and let x repre
sent the distance from it to A, or the distance AC. Then,
c — x will represent the distance B C.
By the conditions of the problem, since the intensity of the
light A, at the distance 1, is a, at the distance x it is — = : and
X'
since the intensity of the light B, at the distance 1, is b, at the
distance c — x it is . ^ . But, by supposition, at C these
intensities are equal ; hence,
a b
Whence,
x 2 ' (c  xf '
c—x
~ ( c 
x) 2
b
or
X 2
a
_ + v*
x \l a
From this equation we obtain as the values of x :
c\J a i \J a
x
or,
\Ja + ^b w \\ja + ^by
c\J a / sj a \
\J a — \J b \\J a — \J b)'
Since both a and b are positive, the two values of x are both
real. Hence,
There are two points of equal illumination on the line of the
lights.
DISCUSSION OF PROBLEMS. 261
Since there are two lights, c must always he greater than ;
consequently neither a, b, nor c can he 0. The problem, then,
admits properly of only these three different suppositions :
1. a > b. 2. a < b. 3. a=b.
We shall now discuss the values of x under each of these
suppositions.
1. a > b.
In this case, the first value of x is less than c ; because
 being a proper fraction, is less than 1. This value
y/ a + \J b
c
of x is also greater than  ; because, the denominator being
less than twice the numerator, as b is less than a, the fraction
is greater than \. Hence, the first point of equal illumination
is at C, between the two lights, but nearer the lesser one.
The second value of x is greater than c : because —A — ,
ya — \J b
being an improper fraction, is greater than 1. Hence, the
second point is at C, in the prolongation of the line A B, be
yond the lesser light.
These results agree with the supposition. For, if a is greater
than b, then B evidently is the lesser light. Hence, both points
of equal illumination will be nearer B than A ; and since the
two lights emit rays in all directions, one of the points must
he in the prolongation of A B beyond both lights.
2. a < b.
In this case, the first value of x is positive. It is also less
than C \ because . ^ C ' . , , having the denominator greater
2 v a + y b
than twice the numerator, b being greater than a, is less than
h Hence, the first point of equal illumination is between the
lights, but nearer A, the lesser light,
The second value of x is negative, because the denominator
y/ a — y 1 b is negative ; which must he interpreted as measur
262 ALGEBKA.
ing distance from A towards the left (Art. 205). Hence, the
second point of equal illumination is at C", in the prolongation
of the line, at the left of the lesser light, A.
These results correspond with the supposition; the case
being the same as the preceding one, except that A is now the
lesser light.
3. a = b.
In this case, the first value of x is positive, and equal to ^ .
Li
Hence, the first point of equal illumination is midway be
tween the two lights.
The second value of x is not finite: because ; — p , if
V « — V b
I
a = b, reduces to ~ = cc (Art. 210), which indicates that no
finite value can be assigned to x. Hence, there is no second
point of equal illumination in the line A B, or its prolongation.
These results agree with the supposition. For, since the
lights are of equal intensity, a point of equal illumination will
obviously be midway between them ; and it is evident that
there can be no other like point in their line.
The preceding discussion illustrates the precision with which
algebraic processes will conform to every allowable interpreta
tion of the enunciation of a problem.
XXXI. — RATIO AND PROPORTION.
337. The Ratio of one quantity to another of the same
kind is the quotient arising from dividing the first quantity
by the second (Art. 181).
Thus, the ratio of a to b is  , or a : b.
b
EATIO AND PROPORTION. 263
338. The Terms of a ratio are the two quantities required
to form it. Of these, the first is called the antecedent, and the
second the consequent.
Thus, in the ratio a : b, a and b are the terms, a the ante
cedent, and b the consequent.
339. A Proportion is an equality of ratios (Art. 181).
Thus, if the ratios a : b and c : d are equal, they form a pro
portion, which may he written
a : b = c : d, or a : b : : c : d.
340. The Terms of a proportion are the four terms of its
two ratios. The first and third terms are called the antece
dents ; the second and fourth, the consequents; the first and
last, the extremes ; the second and third, the means ; and the
terms of each ratio constitute a couplet.
Thus, in a : b = c : d, a and c are antecedents ; b and d, con
sequents ; a and d, extremes; b and c, means; a and b, the
first couplet ; and c and d, the second couplet.
341. A Proportional is any one of the terms of a propor
tion ; a Mean Proportional between two quantities is either
of the two means, when they are equal ; a Third Proportional
to two quantities is the fourth term of a proportion, in which
the first term is the first of the quantities, and the second and
third terms each equal to the second quantity ; a Fourth Pro
portional to three quantities is the fourth term of a proportion
whose first, second, and third terms are the three quantities
taken in their order.
Tims, in a : b = b : e, b is a mean proportional between a
and c ; and c is a third proportional to a and b. In a : b — c : d,
d is a fourth proportional to a, b, and c.
342. A Continued Proportion is one in which each conse
quent is the same as the next antecedent ; as,
a : b = b : c = c : d =■ d : e.
264 ALGEBRA.
PROPERTIES OF PROPORTIONS.
343. Wlien four quantities are in proportion, the prodwt
of the extremes is equal to the product of the means.
Let the proportion be
a : b = c : d.
This may he written (Art. 337),
a c
b = d
Whence, ad = b c.
Hence, if three quantities be in continued proportion, the
product of the extremes is equal to the square of the means.
Thus, if
a : b = b : c
then, a c = b 2 .
By this theorem, if three terms of a proportion are given,
the fourth may be found. Thus, if
then,
Whence,
344. If the product of two quantities be equal to the prod
uct of two others, one pair of them may be made the extremes,
and the other pair the means, of a proportion.
Thus, if
a d = b c
ad be a c
Dividing bv b d, =— =■ = rr— ; , or T = =
& J bd bd' b d
Whence (Art. 337), a:b = c:d.
a : b
= c : x
a x
= bc
x
_bc
a
RATIO AND PROPORTION. 265
In a similar manner, we might derive from the equation
a d = b c, the following proportions :
a : c = b : c 1,
b : d = a : c,
c : d = a : b,
d : b = c : a, etc.
345. If four quantities are in proportion, they will be in
proportion by Alternation; that is, the antecedents will
have to each other the same ratio as the consequents.
Thus, if a : b = c : d
then (Art. 343), ad = bc
Whence (Art. 344), a : c = b : d.
346. If four quantities are in proportion, they will be in
proportion by Inversion; that is, the second term will be to
the first, as the fourth is to the third.
Thus, if a : b — c : d
then, ad — be
Whence, b : a = d : c.
347. If four quantities are in proportion, they will be in
proportion by Composition; that is, the sum of the first two
terms ivill be to the first term, as the sum of the last two terms
is to the third term.
Thus, if a : b — c : d
then, ad = b c
Adding hoth members to a c,
ac + ad = ac + bc, or a (c ' + d) = c {a + b)
Whence, a + b: a = c + d: c (Art. 344).
Similarly, we may show that
a + b : b — c + d : d.
266 ALGEBRA.
348. If four quantities are in proportion, they will be in
proportion by Division; that is,, the difference of the first two
terms will be to the first term, as the difference of the last two
terms is to the third term.
Thus, if a : b = c : d
then, a d = b c
Subtracting both members from a c,
ac — ad = ac — be, or a (c — d)=c (a — b)
Whence, a — b : a = c — d : c.
Similarly, we may prove that
a — b : b = c — d : d.
349. If four quantities are in proportion, they will be in
proportion by Composition and Division; that is, the sum
of the first two terms will be to their difference, as the sum of
the last two terms is to their difference.
(1)
(2)
Thus, if
a : b — c : d
by Art. 347,
a + b c + d
a e
and, by Art. 348,
a — b c — d
a c
Dividing (1) by (2),
a + b c + d
a — b c — d
or, a f b
: a — b = c + d : c
d.
350. Quantifies which are proportional to the same quan
tities, are propjortional to each other.
Thus, if a :b = e:f
and c : d = e :f
. ae.ee
then, r = 3 and  7 = ^
Therefore, =
b d
Whence, a:b = c:d.
RATIO AND PROPORTION. 267
351. If any number of quantities are proportional, any
antecedent is to its consequent, as the sum of all the antece
dents is to the sum of all the consequents.
Thus, if
a : b = c : d = e :f
then (Art. 343), ad = b c
and af=be
also, a b = « b
Adding, a (b + d +/) = b (a + c + e)
Whence (Art. 344), a :b =a + c + e : b + d +f
352. If four quantities are in proportion, if the first and
second be multiplied or divided by any quantity, as also the
third and fourth, the resulting quantities will be in proportion*
Thus, if
a : b = c : d
then,
a c
b = d
Therefore,
ma nc
m b n d
Whence,
m a : mb = n c : n d.
In a similar
manner we could prove
a b c d
m ' m n' n'
Either m or n may be made equal to unity. That is, either
couplet may be multiplied or divided, without multiplying or
dividing the other.
353. If four quantities are in proportion, and the first ami
third be multiplied or divided by any quantity, as also the
second and fourth, the resulting quantities will be in jiro
portion.
268 ALGEBRA.
Thus, if a : b = c : d
then,
Therefore,
a c
b = d
m a m c
nb n d
Whence, m a : nb = m e : nd.
In a similar manner we could prove
a b _ c d
m n m ' n
Either m or n may he made equal to unity.
354. If there be two sets of proportional quantities, the
products of the corresponding terms will be in proportion.
Thus, if a :b = c : d
and e:f=g:h
a c .. e q
then,  =  and =
Therefore,
b d f h
ae eg
bf dh
Whence, ae :bf= c g : d h.
355. If four quantities are in proportion, like poioers or
like roots of these quantities will be in proportion.
Thus, if a :b = c :d
then, r = ,', therefore, — = —
b d ' b n d n
Whence, a n :b n = c n : d n .
In a similar manner we could prove
y/ a : y/ b = y/ c : y d.
356. If three quantities are in continued proportion, the
first is to the third, as the square of the first is f>> the square
of the second.
RATIO AND PROPORTION. 269
Thus, if a:b = b : c
a b
then,
a
b c
a 2 aba
Multiplying by , y=l* c c
Whence, a: c — a~ : b~.
In a similar manner we could prove that if
a : b = b : c = c : d, then a\d=a z : b 3 .
Note. The ratio a 2 : b 2 is called the duplicate ratio, and the ratio a? : b 3
the triplicate ratio, of a : b.
PROBLEMS.
357. 1. The last three terms of a proportion being 18, 6,
and 27, what is the first term ?
2. The first, third, and fourth terms of a proportion being
4, 20, and 55, respectively, what is the second term ?
3. Find a fourth proportional to \, \, and \.
4. Find a third proportional to 5 and 3.
5. Find a mean proportional between 2 and 8.
6. Find a mean proportional between 6 and 24.
7. Find a mean proportional between 49 and 4.
8. Find two numbers in tbe ratio of 2\ to 2, such that when
each is diminished by 5, they shall be in the ratio of lj to 1.
9. Divide 50 into two such parts that the greater increased
by 3, shall be to the less diminished by 3, as 3 to 2.
10. Divide 27 into two such parts that their product shall
be to the sum of their squares as 20 to 41.
11. There are two numbers which are to each other as 4 to
9, and 12 is a mean proportional between them. What are
the numbers ?
270 ALGEBRA.
12. The sum of two numbers is to their difference as 10 to
3, and their product is 364. What are the numbers ?
13. Find two numbers such that if 3 be added to each, they
will be in the ratio of 4 to 3 ; and if 8 be subtracted from each,
they will be in the ratio of 9 to 4.
14. There are two numbers whose product is 96, and the
difference of their cubes is to the cube of their difference as 19
to 1. What are the numbers ?
15. Each of two vessels contains a mixture of wine and
water ; a mixture consisting of equal measures from the two
vessels, contains as much wine as water; and another mixture
consisting of four measures from the first vessel and one from
the second, is composed of wine and water in the ratio of 2 to
3. Find the ratio of wine to water in each vessel.
16. If the increase in the number of male and female
criminals be 1.8 per cent, while the decrease in the number of
males alone is 4.6 per cent, and the increase in the number of
females alone is 9.8 per cent, compare the number of male and
female criminals, respectively, at the first time mentioned.
17. A railway passenger observes that a train passes him,
moving in the opposite direction, in 2 seconds ; whereas, if it
had been moving in the same direction with him, it would
have passed him in 30 seconds. Compare the rates of the two
trains.
XXXII. — VARIATION.
358. Variation, or general proportion, is an abridged
method of expressing common proportion.
Thus, if A and B be two sums of money loaned for equal
times, at the same rate of interest, then
A : B = A's interest : B's interest
VARIATION. 271
or, in an abridged form, by expressing only two terms, the in
terest varies as the principal ; thus (Art. 23),
The interest oc the principal.
359. One quantity varies directly as another, when the
two increase or decrease together in the same ratio.
Sometimes, for the sake of brevity, we say simply one quan
tity varies as another, omitting the word " directly/'
Thus, if a man works for a certain sum per day, the amount
of his wages varies as the number of days during which he
works. For, as the number of days increases or decreases, the
amount of his wages will increase or decrease, and in the same
ratio.
360. One quantity varies inversely as another, when the
first varies as the reciprocal of the second.
Thus, if a courier has a fixed route, the time in which he
will pass over it varies inversely as his speed. That is, if he
double his speed, he will go in half the time; and so on.
361. One quantity varies as two others jointly, when it
has a constant ratio to the product of the other two.
Thus, the wages of a workman will vary as the number of
days he has worked, and the wages per day, jointly.
362. One quantity varies directly as a second and inversely
as a third, when it varies jointly as the second and the recip
rocal of the third.
Thus, in physics, the attraction of a planetary body varies
directly as the quantity of matter, and inversely as the square
of the distance.
363. If A varies as B, then A is equal to B multiplied by
some constant quantity.
Let a and b denote one pair of corresponding values of two
quantities, and A and B any other pair. Then, by Art. 358,
272 ALGEBRA.
A:a = B:b
Whence (Art. 343), Ab = a B, or A=^B
• a
Denoting the constant ratio  by m,
A = m B.
Hence, also, when any quantity varies as another, if any
two pairs of values of the quantities be taken, the four will be
proportional.
For, if A oc B, and. A' and B' be any pair of values of A and
B, and A" and B" any other pair, by Art. 363,
A' = m B', and A" = m B"
a
A 1 A
Whence, ~ = m, and — 
Therefore,
A"
B 1 B"
or (Art. 337), A':B' = A" : B
'ii
364. The terms used in Variation may now be distin
guished as follows :
1. If A = m B, A varies directly as B.
Ml
2. If A = — , A varies inversely as B.
3. If A = m B C, A varies jointly as B and C.
4. If A — , A varies directly as B, and inversely as C.
Problems in variation, in general, arc readily solved by con
verting the variation into an equation, by the aid of Art. 364.
VAKIATION. 273
EXAMPLES.
365. 1. Given that y oc x, and when x = 2, y = 10. Re
quired the value of y in terms of x.
If y oc x, by Art. 364, y = mx
Substituting x = 2 and y = 10, 10 = 2 m, whence m = 5.
Hence, the required value is y = 5 a?.
2. Given that 7/ oc »■, and that v/ = 2 when x = l. What
will he the value of y when x = 2 ?
3. If y oc £, and y = 21 when £ = 3, find the value of y in
terms of z.
4. If x varies inversely as y, and x = 4 when y = 2, what is
the value of cc when y = 6?
5. Given that z varies jointly as x and y, and that z = 1
when x = 1 and ?/ = 1. Find the value of z when x = 2 and
y = 2.
6. If ?/ equals the sum of two quantities, of which one is
constant, and the other varies as x y ; and when x = 2, y = — 2 J,
hut when x = — 2, y = 1 ; express y in terms of x.
7. Two circular plates of gold, each an inch thick, the diam
eters of which are 6 inches and 8 inches, respectively, are
melted and formed into a single circular plate one inch thick
rind its diameter, having given that the area of a circle varies
as the square of its diameter.
8. Given that the illumination from a source of light varies
inversely as the square of the distance ; how much farther from
a candle must a book, which is now 3 inches off, be removed,
so as to receive just half as much light ?
' 9. A locomotive engine without a train can go 24 miles an
hour, and its speed is diminished by a quantity which varies
as the square root of the number of cars attached. With four
cars its speed is 20 miles an hour. Find the greatest number
of cars which the engine can move.
274 ALGEBEA.
XXXIII. — ARITHMETICAL PROGRESSION.
366. An Arithmetical Progression is a series of quanti
ties, in which each term is derived from the preceding term
by adding a constant quantity, called the com raon difference.
367. When the series is increasing, as, for example,
1,3,5,7,9,11,
each term is derived from the preceding term by adding a
positive quantity; consequently the common difference is
positive.
When the series is decreasing, as, for example,
19, 17,15,13,11,9,
each term is derived from the preceding term by adding a
negative quantity; consequently the common difference is
negative.
368. Given the first term, a, the common difference, d, and
the number of terms, n, to find the last term, I.
The progression will be
a, a + d, a + 2 d, a + 3 d,
We observe that these terms differ only in the coefficient of
d, which is 1 in the second term, 2 in the third term, 3 in the
fourth term, etc. Consequently in the rath term, the coefficient
of d will be n — 1. Hence, the rath term of the series, or the
last term, as the number of the terms is n, will be
l = a+ (nl)d (1)
369. Given the first term, a, the last term, 1, and the num
ber of terms, n, to find the sum of the series, S.
S=a+ (a + d) + (a + 2d) + + (l2d) + (l — d) + l
Writing the terms of the second member in the reverse
order,
S=l+ (ld) + (I 2 d) + + {a + 2 d) + (a + d) + a
ARITHMETICAL PROGRESSION. 275
Adding these equations, term by term, we have
2S=(a+l) + (a + l) + (a + l)+ + (a+t)+(a+t) + (a+l)
In this result, (a + I) is taken as many times as there are
terms, or n times ; hence
2S=n(a+l),ovS= 7i (a+I) (2)
Using the value of I given in (1), Art, 3G8, this may he
written
S=%[2a + (nl)d]
370. 1. In the series 5, 8/11, to 18 terms, find the
last term and the sum of the series.
Here a = 5, n = 18 ; the common difference is always found
by subtracting the first term from the second; hence
d = 85 = 3.
Substituting these values in (1) and (2), we have
I = 5 + (18  1) 3 = 5 + 17 x 3 = 5 + 51 = 56.
1 S
S = ^ (5 + 56) = 9 x 61 = 549.
2. In the series 2,1,4, to 27 terms, find the last
term and the sum of the series.
Here a = 2, n = 27, d = the second term minus the first
= — 1 — 2 = — 3. Substituting these values in (1) and (2),
we have
1 = 2 + (271) (3) = 2 + 26 (3) =2 78 = 76.
S=^(276)=^(74) = 27(37) = 999.
«
EXAMPLES.
Find the last term and the sum of the series in the fol
lowing :
276 ALGEBRA.
3. 1, 6, 11, to 15 terms.
4. 7, 3, — 1, to 20 terms.
5. — 9, — 6,3, to 23 terms.
6. 5,10,15, to 29 terms.
^ / o o .
o> 7> k> to 16 terms.
8. ■=, Tp , to 19 terms.
5 15
1 5
9.  , — , to 22 terms.
2 1
10. —  ,  , to 14 terms.
o o
5
11. — 3, — , to 17 terms.
113
12. j j 9 j j 3 to 35 terms.
371. Formulae (1) and (2) constitute two independent
equations, together containing all the five elements of an
arithmetical progression ; hence, when any three of the five
elements are given, we may readily deduce the values of the
other two, as by substituting the three known values we shall
have two equations with only two unknown quantities, which
may be solved by methods previously given.
1. The first term of an arithmetical progression is 3, the
number of terms 20, and the sum of the terms 440. Find the
last term and the common difference.
Here a = 3, n = 20, £=440; substituting in (1) and (2),
we have
1 = 3 + 19 d
 440 = 10 (3 + I), or 44 = 3 + I
From the second equation, I = 41 ; substitute in the first,
41 = 3 + 19 d ; 19 d = 38 ; d = 2.
ARITHMETICAL PROGRESSION. 277
2. Given d = — 3, I = — 39, S = — 264 ; find a and n.
Substituting the given quantities in (1) and (2),
 39 = a + (n  1) ( 3), or 3 n  a = 42
 264 =  (a  39), or a n — 39 n =  528
From the first of these equations, a — 3 ?i — 42 ; substitute
in the second,
(3 n  42) n  39 » =  528, or ?r  27 n =  176
Whence, n =  ~ = ^—^ — = 16 or 11
Substituting in the equation a = 3n — 42,
When ii — 16, a = 6
n = 11, a. = — 9, Ans.
The signification of the two answers is as follows :
If n = 16, and a = 6, the series will be
6, 3, 0,  3,  6,  9,  12,  15,  18,  21,  24,  27,
 30,  33,  36,  39.
If n = 11, and a = — 9, the series will be
_ o,  12,  15,  18,  21,  24,  27,  30,  33,  36,  39.
In either of which the last term is — 39 and the sum — 264.
113
3. Given a = ^ , d — — T x, S = — ^; find I and n.
Substituting the given quantities in (1) and (2), we have
J = +(»_l)(l),or 12l + n*=5
S_nfl
'2~2
(= + l) , ox n + 3 1 n = — 9
278 ALGEBRA.
From, the first of these, n = 5 — 12 I ; substitute in the
second,
5  12 1 + 3 I (5  12 I) =  9, or 36 V"  3 1 = 14
, 3 ±y/ 9 + 2016 3 + 45 2 7
Whence, J = ^ = 70 = 3 ° r ~~ 12
Substituting in the equation re = 5 — 12 £,
9
"When I = o , n — — 3
o
1 = — — , n = 12, Ans.
The first answer is inapplicable, as a negative number of
terms has no meaning. Hence the only answer is,
7
l = — j2> ft = 12.
Note. A negative or fractional value of n is always inapplicable, and
should be neglected, together with all other values dependent on it.
EXAMPLES.
4. Given d = 4, 1 = 75, re = 19 ; find a and >S f .
165
5. Given cZ = — 1, re = 15, $ = —  ; find a and Z.
j
2
6. Given a = — , re = 18, £ = 5 ; find d and &
o
3
7. Given a = — — , re = 7, $ = — 7 ; find d and Z.
8. Given I = — 31, re = 13, S=  169 ; find a and tf.
9. Given a = 3, I = 42f , d = 2£ ; find re and S.
10. Given a = 7, l = — 73, #=— 363 ; find re and c?.
n n 15 5 2625
11. Given a = — , d = 7 :, o= ; find re and Z.
£ Ji Ji
ARITHMETICAL PROGRESSION. 279
12. Given I =  47, d =  1, 8=  1118 ; find a and n.
13. Given d = — 3, S — — 328, a = 2; find Z and n.
372. 2b insert any number of arithmetical means between,
two given terms.
1. Insert 5 arithmetical means between 3 and — 5.
This may he performed in the same manner as the examples
in the previous article ; we have given the first term a = 3 ;
the last term I = — 5 ; the number of terms n = 7 ; as there
are 5 means and two extremes, or in all 7 terms. Substituting
in (1), Art. 368, we have
4
— 5 = 3 + 6 d ; or, 6 d = — 8 ; whence, d = — = .
o
4
Hence the terms are obtained by subtracting  from 3 for
4
the first,  from that result for the second, and so on ; or,
3 5 1 1 l  11 5 Ans
«5j o i o j — L t — o j o" > °> sins.
EXAMPLES.
2. Insert 5 arithmetical means between 2 and 4.
3. Insert 7 arithmetical means between 3 and — 1.
4. Insert 4 arithmetical means between — 1 and — 6.
5. Insert 6 arithmetical means between — 8 and — 4.
6. Insert 4 arithmetical means between — 2 and 6.
7. Insert m arithmetical means between a and b.
PROBLEMS.
373. 1. The 6th term of an arithmetical progression is 19,
and the 14th term is 67. Find the first term.
By Art. 368, the 6th term is a + 5 d, and the 14th term is
a + 13 d. Hence,
280 ALGEBRA.
a+ 5d = 19
a ±13 d = 67
Whence, 8 d = 48, or d = 6
Therefore, a = — 11, Ans.
2. Find four quantities in arithmetical progression, such
that the product of the extremes shall he 45, and the product
of the means 77.
Let a, a + d, a + 2 d, and a ± 3 d he the quantities.
Then, by the conditions, a 2 ± 3 a d = 45 (1)
a 2 ± 3 a d + 2 d 2 = 77 (2)
Subtracting (1) from (2), 2 d 2 = 32
d 2 = 16
t?=±4.
If d = 4, substituting in (1), we have
a 2 + 12a = 45
_ 12±V^Ti4TT80 12 ±18 Q
Whence, a = —^ = ^ = 3 or — 15.
This indicates two answers,
3, 7, 11, and 15, or, — 15, — 11, — 7, and — 3.
If d = — 4, substituting in (1), we have
a 2 — 12 a 45
W1 12±V144 + 180 12 ±18 1K Q
Whence, a = ! — ^ = „ = 15 or — 3.
This also indicates two answers,
15, 11, 7, and 3, or, — 3, — 7, — 11, and — 15.
But these two answers are the same as those obtained with
the other value of d. Hence, the two answers to the problem
are
3, 7, 11, and 15, or, — 3, — 7, — 11, and — 15.
ARITHMETICAL PROGRESSION. 281
3. Find the sum of the odd numbers from 1 to 100.
4. A debt can be discharged in a year by paying $ 1 the
first week, $ 3 the second, $ 5 the third, and so on. .Required
the last payment, and the amount of the debt.
5. A person saves $270 the first year, $210 the second,
and so on. In how many years will a person who saves every
year $ 180 have saved as much as he ?
6. Two persons start together. One travels ten leagues a
day, the other eight leagues the first day, which he augments
daily by half a league. After how many days, and at what
distance from the point of departure, will they come together ?
7. Find four numbers in arithmetical progression, such that
the sum of the first and third shall be 22, and the sum of the
second and fourth 36.
8. The 7th term of an arithmetical progression is 27 ; and
the 13th term is 51. Find the first term.
9. A gentleman set out from Boston to NeW York. He
travelled 25 miles the first day, 20 miles the second day, each
day travelling 5 miles less than on the preceding. How far
was he from Boston at the end of the eleventh day ?
10. If a man travel 20 miles the first day, 15 miles the sec
ond, and so continue to travel 5 miles less each day, how far
will he have advanced on his journey at the end of the 8th
day?
11. The sum of the squares of the extremes of four quanti
ties in arithmetical progression is 200, and the sum of the
squares of the means is 13G. What are the quantities ?
12. After A had travelled for 2 hours, at the rate of 4
miles an hour, B set out to overtake him, and went 4£ miles
the first hour, 4J the second, 5 the third, and so on, increasing
his speed a quarter of a mile every hour. In how many' hours
would he overtake A ?
282 ALGEBRA.
XXXIV. — GEOMETRICAL PROGRESSION.
374. A Geometrical Progression is a series in which each
term is derived from the preceding term by multiplying by a
constant quantity, called the ratio.
375. When the series is increasing, as, for exanrple.
2, 6, 18, 54, 162,
each term is derived from the preceding term by multiplying
by a quantity greater than 1 ; consequently the ratio is a
quantity greater than 1.
When the series is decreasing, as, for example,
9 3 1 i i jl
each term is derived from the preceding term by multiplying
by a quantity less than 1 ; consequently the ratio is a quantity
less than 1.
Negative values of the ratio are admissible ; for example,
3, 6,12,24,48,
is a progression in which the ratio is — 2.
376. Given the first term, a, the ratio, r, and the number
of terms, n, to find the last term, I.
The progression will be
a, ar, ar 2 , ar 5 ,
We observe that the terms differ only in the exponent of r,
which is 1 in the second term, 2 in the third term, 3 in the
fourth term, etc. Consequently in the nth. or last term, the
exponent of r will be n — 1, or
l=ar n  x (1)
377. Given the first term, a, the last term, I, and the ratio.
r, to find the sum of the series, S.
GEOMETRICAL PROGRESSION. 283
S= a + a r + a r 2 + a r 3 + + a r n ~ 3 + a r" 2 + a r n ~ x
Multiplying each term by r,
r S—ar+ ar 2 +ar 3 + a7 A + + a r n ~ 2 + a r" _1 + a r n
Subtracting the first equation from the second, we have
ft 2 ,Tl (f
r S— S=ar n — a, or S (r —l) — a r n — a, or S =  — ^—
But from (1), Art. 376, by multiplying each term by r,
Substituting this value of a r n in the value of S, we have
r — 1
378. 1. In the series 2, 4, 8, to 11 terms, find the last
term and the sum of the series.
Here a = 2, n = ll; the ratio is always found by dividing
4
the second term by the first ; hence, r =  = 2.
Substituting these values in (1) and (2), we have
I = 2 (2) 11  1 = 2 x 2 10 = 2 x 1024 = 2048.
s= (2x204S)2 =4096 _ 2 = 4094
Zi — JL
2. In the series 3, 1. », to 7 terms, find the last term
o
and the sum of the series.
Here a = 3, n = 7, r = second term divided by first term = ^.
Substituting these values in (1) and (2), we have
l  6 \3) ~* \3) _ 3 6_ 3 5_ 243"
284 ALGEBEA.
a j_v , r , 2186
v3 X 243/ ° _729 " 729 _ 2186 3 _ 1093
^ = ~^[ ~  1 — " ^2~ : T29" X 2 ~" ^43" '
3 _1 3 ~3
3. In the series — 2, 6, — 18, to 8 terms, find the last
term and the sum of the series.
Here a — — 2, n = S, r= — ~ = — 3. Hence,
I = (_ 2) ( 3) 8  1 = ( 2) ( 3) 7 = ( 2) ( 2187) = 4374.
(_ 3 x 4374)  ( 2)  13122 + 2 _^  13120 =
(3) — 1 4 4
EXAMPLES.
Find the last term and the sum of the series in the fol
lowing :
4. 1, 2, 4, to 12 terms.
4
5. 3, 2,  , to 7 terms.
o
6. —2, 8, —32, to 6 terms.
7. 2, — 1, ■= , to 10 terms.
111
2' V 8
8. ^ , T , q , to 11 terms.
2 3
9. k i — 1? ^ i to 8 terms.
10. 8, 4, 2, to 9 terms.
11. 4' 4' J2' to6 termS *
2 11
12. k>— g>— gj to 10 terms.
13. 3, 6, 12, to 7 terms.
GEOMETRICAL PROGRESSION. 285
379. Formulae (1) and (2) together contain all the five ele
ments of a geometrical progression ; hence, if any three of the
five are given, we may find the other two, exactly as in arith
metical progression. But in certain cases the operation in
volves the solution of an equation of a higher degree than the
second, for which rules have not heen given ; and in other
cases the unknown quantity appears as an exponent, the solu
tion of which equations can usually only be effected by the use
of logarithms ; although in certain simple cases they may be
solved by inspection.
1. Given I = 6561, r — 3, n = 9 ; find a and S.
Substituting these values in (1) and (2), Arts. 376 and 377,
we have
6561 = a (3) 8 ; or 6561 = 6561 a ; or a = 1.
(3x6561)l_ 196831 _ 19682 _
6 ~ 3=1 ~" ~2~ "" ~2~
2. Given a = — 2, n = 5, I = — 32 ; find r and S.
Substituting these values in (1) and (2), we have
32 = (2)(r) 5  1 ; or32 = 2r 1 ; r 4 = 16; r = ±2.
If r = 2, S= (2 X ~ 32) 7 ( ~ 2) = 64 + 2 = 62.
T , 9 q (2x32)(2) _ 64 + 2 _66_
The signification of the two answers is as follows :
If r=2, the series will be 2, 4, 8, 16, 32, in
which the sum is — 62.
If r = — 2, the series will be — 2, 4, — 8, 16, — 32, in which
the sum is — 22.
3. Given a = 3, r = — ^ , S = ; find n and I.
286 ALGEBRA.
Substituting these values in (1) and (2), we have
Us
1640 _ 3 1+9 • 6560 1
"729— ^TV=^r ; oW+9 = T29 ; ° Tl = 729
3
3
Substituting this value of I in the equation (— 3)" _1 = , we
3
have (— 3) n ~ 1 = ^ = — 2187; whence, by inspection, n— 1
71".)
= 7, or n = 8.
EXAMPLES.
4. Given J =  256, r = — 2, n = 10 ; find a and A
5. Given r = , n = 8, S— ^^r ; find a and Z.
o 6561
6. Given a = 2, n = 7, I = 1458 ; find r and 5.
3
7. Given a = 3, « = 6, Z = — . ,^ , ; find r and &
1024
8. Given a= 1, r = 3, Z = 81 ; find « and S.
1 127
9. Given a = 2, Z = ^ , $ = ^ ; find n and r.
10. Given «. =  , r = — 3, $= — 91; find n and Z.
11. Given £ = 128, r = 2, £ = 255; find n and a.
380. The Limit to which the sum of the terms of a decreas
ing geometrical progression approaches, as the number of terms
becomes larger and larger, is called the sum of the scries to
infinity. We may write the value of S obtained in Art. .'177
as follows :
GEOMETRICAL PROGRESSION. 287
a — r I
S.
1r
In a decreasing geometrical progression, the larger the num
ber of terms taken the smaller will be the value of the last
term ; hence, by taking terms enough, the last term may be
made as small as we please. Then (Art. 207), the limiting
value of I is 0. Consequently the limit to which the value of
S approaches, as the number of terms becomes larger and
larger, is  .
Therefore the sum of a decreasing geometrical progression
to infinity is given by the formula
1. Find the sum of the series 3, 1, ^ , to infinity.
o
Here a = 3, r = k ; substituting in (3), we have
O 3 9 9 A
3
8 16
2. Find the sum of the series 4, — ^ , — , to infinity.
_8
~~ 3 2
Here a = 4, r = —r = — » ; substituting in (3), we have
+ 3
EXAMPLES.
Find the sum of the following to infinity :
3l 2 ' lj 2' 5   1 '3'~9'
288
ALGEBRA.
3 11
'■ 4' 2' 3' '■"
9 8 2 1
9  *'5' 50'
8 2 A li in 4 4 _A
5' 35' 245' ' 5' " 25'
381. To find the value of a rejoeating decimal.
This is a case of finding the sum of a geometrical progres
sion to infinity, and may be solved by the formula of Art. 380.
1. Find the value of .363636
.363636 = .36 + .0036 + .000036 +
AAO/>
Here a = .36, and r = '—— — = .01 ; substituting in (3),
.oo
.36 _ : 36_36_ £
2. Find the value of .285151
.285151 = .28 + .0051 + .000051 +
To find the sum of all the terms except the first, we have
a = .0051, r = .01 ; substituting in (3),
a .0051 .0051 51 17
o =
1.01 .99 9900 3300
Adding the first term to this, the value of the given decimal
28 17 941
~ 100 + 3300" 3300'
EXAMPLES.
Find the values of the following :
3. .074074 5. .7333 7. .113333.
4. .481481 6. .52121 8. .215454.
GEOMETRICAL PROGRESSION. 289
382. To insert any number of geometrical means between
two given terms.
64
1. Insert 4 geometrical means between 2 and pr^ .
This may be performed in the same manner as the examples
64
in Art. 379. We have a = 2, I = —^ , and n = 6, or two more
243
than the number of means.
Substituting these values in (1), Art. 376, we have
64 32 2
243 = 2? ' 5; ° rr5 = 243 ; mr = t
2
Hence the terms are obtained by multiplying 2 by ^ for the
2
first, that result by ^ for the second, and so on ; or,
o
4 8 16 32 ^4
2 '3' 9' 27' 81' 243'
2. Insert 5 geometrical means between — 2 and — 128.
Here a = — 2, I = — 128, n = 7. Substituting in (1), Art.
376, we have
— 128 = — 2 7* ; or r 6 = 64 ; whence, r = ± 2.
If r = 2, the series will be
_ 2; 4, 8, 16, 32, 64, 128.
If r = — 2, the series will be
 2, 4,  8, 16,  32, 64,  128.
EXAMPLES.
o 1 128
3. Insert 6 geometrical means between o and ^Hq •
4. Insert 5 geometrical means between ^ and 364£.
5. Insert 6 geometrical means between — 2 and — 4374.
200 ALGEBRA.
729
6. Insert 4 geometrical means between 3 and —
3
1024
7. Insert 7 geometrical means between  and
PROBLEMS.
383. 1. What is the first term of a geometrical progression,
when the 5th term is 48, and the 8th term is — 384 ?
By Art. 376, the 5th term is a r 4 , and the 8th term is a r~.
Hence,
ar* = 48, and a r' = — 384.
Dividing the second of these equations by the first,
r 3 = — 8 ; whence, r = — 2.
Tl 48 48 <* J
Ihen, « = — ;  = :r7r= z: 3 , Ans.
r* 16
2. Find three numbers in geometrical progression, such that
their sum shall be 14, and the sum of their squares 84.
Let "a, a r, and a r~ denote the numbers. Then, by the con
ditions,
a + a r + a r 2 = 14 (1)
a 2 + a 2 r 2 + a 2 r i = 8A (2)
Dividing (2) by (1), a — ar+ar 2 = 6 (3)
Adding (1) and (3), a + a r 2 = 10 (4)
4
Subtracting (3) from (1), a r = 4, ofr =  (5)
1 c
Substituting from (5) in (4), a \ = 10
<r
10 a = 16
wi (K onox lOiy/10064 10 ±6 Q _
VV hence (Art. 309), a — ~  = —  — — 8 or 2.
— Z
4 1
If a = 8, r = q = jr, and the numbers are 8, 4, and 2.
8 Z
HARMONICAL PROGRESSION. 291
4
If a = 2, ?• =  = 2, and the numbers are 2, 4, and 8.
Therefore, the numbers are 2, 4, and 8, Ans.
3. A person who saved every year half as much again as he
saved the previous year, had in seven years saved $2059.
How much did he save the first year ?
4. A gentleman boarded 9 days, paying 3 cents for the first
day, 9 cents for the second day, 27 cents for the third day, and
so on. Required the cost.
5. Suppose the elastic power of a ball that falls from a
height of 100 feet, to be such as to cause it to rise 0.9375 of
the height from which it fell, and to continue in this way
diminishing the height to which it will rise, in geometrical
progression, till it comes to rest. How far will it have moved ?
6. The sum of the first and second of four quantities in
geometrical progression is 15, and the sum of the third and
fourth is 60. Required the quantities.
7. The fifth term of a geometrical progression is — 324, and
the 9th term is — 26244. What is the first term ?
8. The third term of a geometrical progression is ^r , and
9
the sixth term is ^r^ . What is the second term ?
XXXV.— HARMONICAL PROGRESSION.
384. Quantities are said to be in Harmonical Progression
when their reciprocals form an arithmetical progression.
For example, 1, , F , ,
3' 5' 7
are in harmonical progression, because their reciprocals,
1, 3, 5, 7,
form an arithmetical progression.
292 ALGEBRA.
385. From the preceding it follows that all problems in
harmonical progression, which are susceptible of solution, may
be solved by inverting the terms and applying the rules of the
arithmetical progression. There will be found, however, no
general expression for the sum of a harmonical series.
386. To find the last term of a given ha rmonical series.
2 2
3' 5 ;
1. In the series 2,  , p, to 36 terms, find the last term.
Inverting the series, we have the arithmetical progression
2 j 2 ' 2 ' t0 36 terms •
Here a = , d = l, w = 36 ; hence, by (1), Art. 368,
Z = i+(36l)l = ^ + 35 = ^.
2
Inverting this, we obtain =r as the last term of the given series.
EXAMPLES.
Find the last terms of the following :
K Q A *} 1 9
2.  ,  , to 23 terms. 4. r,T,, to 26 terms.
3.  r, to 17 terms. 5. a, b, to n terms.
2 3 o
387. To ii/si'rf any number of harmonical means between
two given terms.
1. Insert 5 harmonical means between 2 and — 3.
Inverting, we have to insert 5 arithmetical means between
1 A 1
2 and ~3'
HARMONICAL PROGRESSION. 293
Here a =  , Z = — ,n = J; substituting in (1), Art, 368,
Z o
we have
—  —  + 6d: or 6 d — — x : whence, cZ = — — .
3 2 6 ob
Hence, the arithmetical means are
13 2 1_ _1_ _L
36' 9' 12' 18' 36*
Then, the harmonical means will be
36 9 ._ 1Q 36
,,12,lS,  T ,Ans.
EXAMPLES.
2 3
2. Insert 7 harmonical means between  and — .
5 10
3. Insert 3 harmonical means between — 1 and — 5.
4. Insert 6 harmonical means between 3 and — 1.
5. Insert m harmonical means between a and b.
388. If three consecutive terms of a harmonical progres
sion be taken, the first has the same ratio to the third, that
the first minus the second has to the second minus the third.
Let a, b, c be in harmonical progression ; then their recip
rocals  ,  , and  will be in arithmetical progression. Hence^
a> o c
1_1_1_1
c b b a'
Clearing of fractions, ab — ac = ac — bc
or, a (b — c) = c (a — b)
Dividing through by c (b — c), we have
a a — b
which was to be proved.
294 » ALGEBRA.
389. Let a and c be any two quantities ; b their harrnoni
cal mean. Then, by the previous theorem,  = .
Clearing of fractions, ab — ac = ac — be; then, ab + bc = 2ac
2 a c
or, b = .
a + c
390. We may note the following results : if a and c arc
a ~f~ c
any two quantities, their arithmetical mean = —  — ; their
geometrical mean = \a c ; and their harmonical mean =   .
a+ c
a 2 a c a + c / / — \2 .
Since X — 7. — = vV a c ) > l t follows that the product
a \ c ~j
of the harmonical and arithmetical means of two quantities is
equal to the square of their geometrical mean.
Consequently the geometrical mean must be intermediate in
value between the harmonical and the arithmetical mean. But
the harmonical mean is less than the arithmetical mean, be
a + c 2ac (a + e) 2 — 4 a c a 2 + 2 ac+ c 2 — £ac
cause — pr =  — T —/ r = — — r
2 a + c 2 (a + c) 2 (a + c)
a 2 — 2ac + c 2 (a — c) 2
= — 7T7 ^ = 7T^ n a positive quantity.
2 (a + c) 2 (a + c) ' L >■ *
Hence of the three quantities, the arithmetical mean is the
greatest, the geometrical mean next, and the harmonical mean
the least.
XXXVI. — PERMUTATIONS AND COMBINA
TIONS.
391. The different orders in which quantities can be ar
ranged are called their Permutations.
Thus, the permutations of the quantities a, b, <; taken two
at a time, are
PERMUTATIONS AND COMBINATIONS. 295
a b, b a ; a c, c a; be, c b;
and taken three at a time, are
a b c, ac b; b a e, b c a ; cab, cb a.
392. The Combinations of quantities are the different col
lections that can he formed out of them, without regard to the
order in which they are placed.
Thus, the combinations of the quantities a, b, c, taken two
at a time, are
a b, ac, be;
a b, and b a, though different permutations, forming the same
combination.
393. To find the number of permutations of n quantities,
taken r at a time.
Let P denote the number of permutations of n quantities,
taken r at a time. By placing before each of these the other
n — r quantities one at' a time, we shall evidently form P (n — r)
permutations of the n quantities, taken r + 1 at a time. That
is, the number of permutations of n quantities, taken r at a
time, multiplied by n — r, gives the number of j)ermutations
of the n quantities, taken r + 1 at a time.
But the number of permutations of n quantities, taken one
at a time, is obviously n. Hence, the number of permutations
taken two at a time, is the number taken one at a time,
multiplied by n — 1, or n (n— 1). The number of permuta
tions, taken three at a time, is the number taken two at a time,
multiplied by n — 2, or n (n — 1) (n — 2); and so on. We
observe that the last factor in the number of permutations is
n, minus a number 1 less than the number of quantities taken
at a tinie. Hence, the number of permutations of n quanti
ties, taken r at a time, is
n(nl) 02) (n(rl))
or, n(n — l)(n — 2) (n — r + T). (1)
296 ALGEBRA.
394. If all the quantities are taken together, r = n and
Formula (1) becomes
n(nl) (n2) 1;
or, by inverting the order of the factors,
1x2x3 (n— l)n. (2)
That is,
The number of permutations of n quantities, taken n at a
time, is equal to the product of the natural numbers from 1
up to n.
For the sake of brevity, this result is often denoted by \n,
read "factorial n " ; thus \n_ denotes the product of the natu
ral numbers from 1 to n inclusive.
395. To find the number of combinations of n quantities,
taken r at a time.
The number of permutations of n quantities, taken r at a
time, is (Art. 393),
n{n — l) (n — 2) (n — r+ 1).
By Art. 394, each combination of r quantities produces \r
permutations. Hence, the number of combinations must equal
the number of permutations divided by ta or
n(nl) (n2) (nr+1)
 r
396. The number of combinations of n quant It Its, taken r
at a time is the same as the number of combinations of n
quantities taken n — r at a time.
For, it is evident that for every combination of r quantities
which we take out of n quantities, we leave one combination
of n — r quantities, which contains the remaining quantities.
EXAMPLES.
397. 1. How many changes can be rung with 10 bells,
taking 7 at a time ?
PERMUTATIONS AND COMBINATIONS. 2! >7
Here, n = 10, r = 7 ; then n — r + 1 = 4.
Then, by Formula (1),
10x9x8x7x6x5x4 = 604800, ^».s.
2. How many different combinations can be made with 5
letters out of 8 ?
Here, n = 8, r = 5 ; then n — r + 1 = 4:.
Then, by Formula (3),
8x7x6x5x4
1x2x3x4x5
56, Ans.
3. In how many different orders may 7 persons be seated
at table ?
Here n = 7 ; then, by Formula (2),
1x2x3x4x5x6x7 = 5040, Ans.
4. How many different words of 4 letters each can be made
with 6 letters ? How many words of 3 letters each ? How
many of 6 letters each ? How many in all possible ways ?
5. How often can 4 students change their places in a class
of 8, so as not to preserve the same order ?
6. From a company of 40 soldiers, how many different pick
ets of 6 men can be taken ?
7. How many permutations can be formed of the 26 letters
of the alphabet, taken 4 at a time ?
8. How many different numbers can be formed with the
digits 1, 2, 3, 4, 5, 6, 7, 8, 9, taking 5 at a time, each digit
occurring not more than once in any number ?
9. How many different permutations may be formed of the
letters in the word since, taken all together ?
10. How many different combinations may be formed of the
letters in the word forming, taken three at a time ?
298 ALGEBRA.
11. How many different combinations may be formed of 20
letters, taken 5 at a time ?
12. How many different combinations may be formed of 18
letters, taken 11 at a time ?
13. How many different committees, consisting of 7 persons
each, can be formed out of a corporation of 20 persons ?
14. How many different numbers, of three different figures
each, can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, ?
XXXVII. — BINOMIAL THEOREM.
POSITIVE INTEGRAL EXPONENT.
398. The Binomial Theorem, discovered by Newton, is a
formula, by means of which any binomial may be raised to any
required power, without going through the process of invo
lution.
399. To prove the Theorem for a positive integral ex
ponent.
By actual multiplication we may show that
(a + x) 2 = a 2 + 2 ax + x 2
(a + x) 3 = a 3 + 3 a 2 x + 3 a x 2 + x 3
(a + x) 4 = a 4 + 4 a 3 x + 6 a 2 x 2 + 4 a x 3 + x*
In these results we observe the following laws :
1. The number of terms is one more than the exponent of the
binomial.
2. The exponent of a in the first term is the same as the
exponent of the binomial, and decreases by one in each stic
ceedin;/ term.
3. The exponent of x in the second term is unity, and in
creases by one in each succeeding term.
BINOMIAL THEOREM. 299
4. The coefficient of the first term, is unity ; and of the sec
ond term, is the exponent of the binomial.
5. If the coefficient of any term be multiplied by the expo
nent of a in that term, and the product divided by the number
of the term, beginning at the left, the result will be the co
efficient of the next term,.
Assuming that the laws hold for any positive integral expo
nent, n r we have
. . , n(n— 1) „ „ n(n— T)(n— 2) „ ,
(a+x) n =a n +na n  1 x+ \ ' a n ~ 2 x 2 + K J ± / a n ~ s x 3 +
This result is called the Binomial Theorem.
400. To prove that it holds for any positive integral expo
nent, we multiply hoth members by a + x, thus
/ x ^i , , n(n—T) , , n(n — l)(n— 2) . „
(a+x) n+1 =a n+1 +?ia n x + \ V ^ar^ v J \ ' a n ~ 2 x 3
, „ n(n— 1) . ,
+ + a n x + na n ~ y x~+ v a n ~ i x 6 +.
X.A
= a n + 1 + (n + 1) a n x +
n (n — 1)
a"" 1 x
ln{nl)(n2) ^1) 1
+ L 1.2.3 " + 1.2 J +>
= a n+1 + (w + 1) a n x + ™ 2 In  1 + 2 1 a n ~ l a?
+ n S [H <*+
{ ?i "T~ 1 ) 72
= a" +1 + (n + 1) a" a; + ^jtj a" 1 ar 2
(w + 1) n (n— 1) _„ 3
1.2.3 ~ a " " * +
where it is evident that every term except the first will con
tain the factor n + 1.
300 ALGEBRA.
"We observe that the expansion is of the same form as the
value of (a + x) n , having n + 1 in the place of n.
Hence, if the laws of Art. 399 hold for any positive integral
exponent, n, they also hold when that exponent is increased
by 1. But we have shown them to bold for (a + cc) 4 , hence
they hold for (a + x) 5 ; and since they hold for (a + x) 5 , they
also hold for (a + x) 6 ; and so on. Hence they hold for any
positive integral exponent.
401. Since 1.2 = [2, 1. 2. 3 = [3, etc. (Art. 394), the Bino
mial Theorem is usually written as follows :
\ n(n—l) „ o n(n—l)(n—2) _ , „
(a+x) n =a n +?i a n ~ l x+ V V ~V+— £ } a n ^x 3 +
If. £L
402. If a =1, then, since any power of 1 equals 1, we
bave
?i(n — l) , w (n. — 1) (w — 2) „
403. In performing examples by the aid of the Binomial
Theorem, we may use the laws of Art. 399 to find the expo
nents and coefficients of the terms.
1. Expand (a + x) 6 by the Binomial Theorem.
The number of terms is 7.
The exponent of a in the first term is 6, and decreases by 1
in each succeeding term.
The exponent of x in the second term is 1, and increases by
1 in each succeeding term.
The coefficient of the first term is 1 ; of the second term, 6;
if the coefficient of the second term, 6, be multiplied by 5, the
exponent of a in that term, and the product, 30, be divided by
the number of the term, 2, the result, 15, will be the coefficient
of the third term ; etc.
Eesult, a 6 +6a 5 x + 15 a 4 x 2 + 20 a 3 x* + 15 a 2 x" + 6 a x 5 + x*.
BINOMIAL THEOREM. 301
Note. It will be observed that the coefficients of any two terms taken
equidistant from the beginning and end of the expansion are the same.
The reason for this will be obvious if, in Art. 401, x and a be inter
changed, which is equivalent to inverting the series in the second member.
Thus, the coefficients of the latter half of an expansion may bo written
out from the first half.
2. Expand (1 + sc) 7 by the Binomial Theorem. "
Result, l 7 + 7.1 6 . x + 21.1 5 . x 2 + 35.1 4 . x 3 + 35.1 3 . x* + 21.1 s . x 5
+ 7.1 1 . x° + x 1 ;
or, 1 + 7 x + 21 x 2 + 35 x 3 + 35 x A + 21 x 5 + 7 x 6 + x\
Note. If the first term of the binomial is a numerical quantity, it will
be found convenient, in applying the laws, to retain the exponents at first
without reduction, as then the laws for coefficients may be used. The re
sult should afterwards be reduced to its simplest form.
3. Expand (2 a + 3 b)'° by the Binomial Theorem.
(2a + 3&) 5 =[(2«.) + (36)] 5
= (2 a) 5 + 5 (2 a) 4 (3 b) + 10 (2 a) 3 (3 bf + 10 (2 a)' 2 (3 b) 3
+ 5 (2 a) (3 by + (3 b) 5
= 32 a 5 + 240 a 4 b + 720 a 3 V 2 + 1080 a 2 b 3 + 810 a b* + 243 b 5 ,
Aiis.
4. Expand (irf 2 — ft 1 ) 6 by the Binomial Theorem.
( m i _ n i)« = [(m _i ) + ( ft 1 )] 6
= ( m ^) +6(m"2)5(_ w i) + i5( m *) 4 (« 1 )' 2 +20(m"^) 3 (H 1 ) 3
+ 15 (m~ty (ft" 1 ) 4 + G (m~*) (n l f+ (ft 1 ) 6 '
=mr 3 + 6 m~ r% ( ft" 1 ) + 15 m~ 2 (ft 2 ) + 20 i»~* ( ft" 3 )
+ 15 m 1 (ft 4 ) + 6 m~ ( ft" 5 ) + (ft" 6 )
= m~ 3 — 6 m~* ft 1 + 15 m~ n~ — 20 m~  ft~ 3 + 15 m~ x ft" 4
— 6 m " 2 ft 5 + ft 6 , Ans.
302 ALGEBRA.
Note. If either term of the binomial is not a single letter, with unity
as its coefficient and exponent, or if either term is preceded by a minus
sign, it will be found convenient to enclose the term, sign and all, in a
parenthesis, when the usual laws for exponents and coefficients may be
applied. In reducing, care must be taken to apply the principles of Arts.
227 and 259.
EXAMPLES.
Expand the following by the Binomial Theorem :
5.
(1 + o) 5 .
8.
(a b — c d)~.
11.
(J + <$y.
6.
(a + a 3 ) 6 .
9.
O 2 + 3 nr)\
12.
(m~% + 2 ?i s y.
7.
(x*2yy.
10.
(a 2  4 x*) b .
13.
(ai — &»aj4)«.
404. To find the rth or general term of the expansion of
(a + ,r)\
"We have now to determine, from the observed laws of the
expansion, three things ; the exponent of a in the term, the
exponent of x in the term, and the coefficient of the term.
The exponent of x in the second term is 1 ; in the third
term, 2 ; etc. Hence, in the rth term it will be r — 1.
In any term the sum of the exponents of a and x is n.
Hence, in the rth term, the exponent of a will be such a quan
tity as when added to r — 1, the exponent of x, will produce
n ; or, the exponent of a will be n — r + 1.
The coefficient of the term will be a fraction, of the form
n(nl)(»2) , . , , . , ,
. —  } — q— ; m which we must determine the last
X . w . o
factors of the numerator and denominator.
We observe that the last factor of the numerator of any
tennis 1 more than the exponent of a in that term: hence
the last factor of the numerator of the rth term will be
n — r + 2.
Also, the last factor of the denominator of any term is the
same as the exponent of x in that term : hence the last factor
of the denominator of the rth term will be r — 1.
BINOMIAL THEOREM. 303
Therefore the
n (n  1) (n  2) (n — r + 2)
rth term = ^ t^t;^ — ^^^rr  1 — ct n ~ r + 1 x r ~\
1 . Z . 6 (r — 1)
1. Find the 8th term of (3 J 2b~ l ) n .
Here r = 8, n = 11 ; hence, the
8thtom = 1 l:2°3 9 4 8 5.6 6 7 5 < 3a *> , ( 2t )'
= 330 (81 a 2 ) ( 128 b~'') =  3421440 a 2 J 7 , ^ras.
Note. The note to Ex. 4, Art. 403, applies with equal force to examples
in this article.
EXAMPLES.
Find the
2. 10th term of (a + x) 15 . 5. 5th term of (1  a 2 ) 12 .
3. 6th term of (1 + m) u . 6. 9th term of (x~ 1 2 y*) u .
4. 8th term of (c  d) l \ 7. 8th term of (a% + 3 x" 1 ) 10 .
405. A trinomial may he raised to any power by the Bi
nomial Theorem, if two of its terms he enclosed in a paren
thesis and regarded as a single term ; the operations indicated
being performed after the expansion by the Theorem has been
effected.
1. Expand (2 a — b + c 2 ) 3 by the Binomial Theorem.
(2 a  b + c 2 ) 3 = [(2 ab) + (c 2 )] 3
= (2 a  by + 3 (2 a b) 2 (c 2 ) + 3 (2 a  b) (c 2 ) 2 + (c 2 ) 3
=Sa 3 ~12a 2 b + 6ab 2 b 3 +3c 2 (4:a 2 4ab + b 2 ) + 3c 4 (2ab) + c 6
= 8 a 3  12 a 2 b + 6 a b 2  b 3 + 12 a 2 c 2  12 a b c 2 + 3 b 2 c 2
+ 6ac i 3bc 4 + c 6 , Ans.
304 ALGEBRA.
The same method will apply to the expansion of any poly
nomial hy the Binomial Theorem.
EXAMPLES.
Expand the following by the Binomial Theorem :
2. (1xx 2 )*. 4. (1  2 x  2 x 2 )\
3. (x+3x + l) 3 . 5. (l + a;a; 2 ) 5 .
XXXVIII.— UNDETERMINED COEFFICIENTS.
406. A Series is a succession of terms, so related that each
may be derived from one or more of the others, in accordance
with some fixed law.
The simpler forms of series have already been exhibited in
the progressions.
407. A Finite Series is one having a finite number of
terms.
408. An Infinite Series is one whose number of terms is
unlimited.
The progressions, in general, are examples of finite series ;
but, in Art. 380, we considered infinite Geometrical series.
409. An infinite series is said to be convergent when the
sum of the first n terms cannot numerically exceed some finite
quantity, however large n may be ; and it is said to be diver
gent when the sum of the first n terms can numerically exceed
any finite quantity by taking n large enough.
For example, consider the infinite series
1 + x + x + X 3 +
The sum of the first n terms
1 + x + x 2 + x 3 + +x n ~ 1 = ^^ (Art. 120).
1 — x
UNDETERMINED COEFFICIENTS. 305
If x is less than 1, x n is less than x, however largo n may
be ; consequently the numerator and denominator of the frac
tion are each less than 1, and positive ; and the numerator is
larger than the denominator ; hence the fraction is equal to
some finite quantity greater than 1. The series is therefore
convergent if x is less than 1.
If x is equal to 1, each term of the series equals 1, conse
quently the sum of the first n terms is n ; and this can numer
ically exceed any finite quantity by taking n large enough.
The series is therefore divergent if x = 1.
If x is greater than 1, each term of the series after the first
is greater than 1, consequently the sum of the first n terms is
greater than n ; and this sum can numerically exceed any finite
quantity by taking n large enough. The series is therefore
divergent if x is greater than 1.
410. Every infinite literal series, arranged in order of pow
ers of some letter, is convergent for some values of that letter,
and divergent for other values.
We will now show that it is convergent when that letter
equals zero.
Let the series be
a + bx + cx 2 + dx s + + Jex n ~ 1 +
The sum of the first n terms is
a + bx + cx 2 + dx 3 + + kx n ~\
which is equal to a, if x is made equal to 0.
Hence, however large n may be, the sum of the first n terms
is equal to a, if x is equal to 0. Consequently the series is
convergent if x = 0.
411. Infinite series may be developed by the common pro
cesses of Division, as in Art. 101, Exs. 19 and 20, and Extrac
tion of Roots, as in Arts. 239 and 243 ; and by other methods
which it will now be our object to elucidate.
306 ALGEBRA.
UNDETERMINED COEFFICIENTS.
412. A method of expanding algebraic expressions into
series, simple in its principles, and general in its application,
is based on the following theorem, known as the
THEOREM OF "UNDETERMINED COEFFICIENTS.
413. If the series A + Bx + Cx 2 + Bx 3 + is always
equal to the series A' + B' x f C ar f D' x 3 + , for a ny
value of x which makes both series convergent, the coefficients
of like powers of ' x in the two series will he equal.
For, since the equation
A + Bx + Cx 2 +Dx 3 + =A'+B'x+ C'x+D>x 3 +
is satisfied for any value of x which makes both members con
vergent ; and since by Art. 410, if x is equal to 0, both mem
bers are convergent ; it follows that the equation is satisfied if
x = 0. Making x — 0, the equation becomes
A = A'.
Subtracting A from the first member of the equation, and its
equal, A', from the second member, we have
Bx+ Cx 2 +Bx 3 + = B'x + C'x 2 + D'x 3 +
Dividing through by x,
B+Cx + Bx 2 + =B'+ C'x + D'x 2 +
This equation is also satisfied for any value of X which makes
both members convergent ; hence it is satisfied if x = 0. Mak
ing x = 0, we have;
B = B'.
Proceeding in this way, we may show C= O, I) = D', etc.
UNDETERMINED COEFFICIENTS. 307
Note. The necessity for the limitation of the theorem to values of x
which make both series convergent, is that a convergent series evidently
cannot be equal to a divergent series ; and two divergent scries cannot be
equal, as two quantities which numerically exceed any finite quantity can
not be said to be equal.
Hence, in all applications of the theorem, the results are only true when
both members are convergent.
APPLICATION TO THE EXPANSION OF FRACTIONS INTO SERIES.
2 + 5 x
414. 1. Expand  —  into a series.
JL — o x
We have seen (Art. 101), that any fraction may he expanded
into a series by dividing the numerator by the denominator ;
consequently, we know that the proposed development is pos
sible. Assume then,
^ + f x =A+Bx + Cx 2 + Dx*+Ex i + (1)
1 — ox
where A, B, C, D, E, are quantities independent of x.
Clearing of fractions, and collecting together in the second
member the terms containing like powers of x, we have
2+5x = A + B
3.4
x+ C
3B
x+ D
3(7
x 3 + E
3D
x* +
Equation (1), and also the preceding equation, are evidently
to be satisfied by all values of x which make the second mem
ber a convergent series. Hence, applying the Theorem of Un
determined Coefficients to the latter, we have
A = 2.
B — 3 A — 5 ; whence, B = 3 A + 5 = 11.
C3B = 0; whence, C = 3B =33.
D  3 C = ; whence, D = 3 C = 99.
E3D = 0; whence, E=3D =297.
303
ALGEBRA.
Substituting these values of A, B, C, D, E, in (1), we
have
^+^ = 2 + 11 a: + 33 z 2 + 99 cc 3 + 297 a: 4 + ,
1 — o x
which may he readily verified by division.
This result, in accordance with the last part of the Note to
Art. 413, only expresses the value of the fraction for such
values of x as make the second member a convergent series.
2. Expand — — « into a series.
l2ica;' 2
Assume
1 — 3 x
■x*
A + Bx + Cx 2 +Dx !i + JEx i +
l—2x—x
Clearing of fractions, and collecting terms,
1Sx
x 2 = A + B
2 A
x+ C
2B
 A
x 2 + D
2(7
 B
x 3 + E
2D
 C
.v
Equating the coefficients of like powers of x,
A = l.
B2A = 3; whence, B — 2A — 3 = — 1.
C2BA = 1; whence, C=2B + Al = 2.
D2C—B= 0; whence, D=2 C+B = — 5.
E2D 0— 0; whence, E=2D+ C=12.
Substituting these values,
13
x
■x
\2x
■X"
r = 1 — x — 2 x 2 — 5 x 3 — 12 x* — , A ns.
Note. This method enables us to find the law of the coefficients in any
expansion. For instance, in Example 1, we obtained the equations C=3B,
D = BC, E = ZD, etc. ; or, in general, any coefficient, after the second, is
three times the preceding. In Example 2, we obtained the equations
Ij — IC^B, E=2D + C, etc.; or, in general, any coefficient, after the
third, is twice the preceding plus the next but one preceding. After the
law of the coefficients of any expansion has been found, we may write out
the subsequent terms to any desired extent by its aid.
23
x + Ax 2
1 + 2
x — 5x 2 '
3 + x
2x 2
3 — x
+ x 2 '
1
3x 2
UNDETERMINED COEFFICIENTS. 309
EXAMPLES.
Expand the following to five terms :
3. * = ". 6. \~'* . 9.
1 + X 1 + x + x z
3+1* 7 l^ 2 1Q
l5x' l + 2x3a; 2 '
2  a; + x 2 8 l + 2.x n
Sl 1x 2 ' ' 2cca; 2 ' " 2 3 a: 2 a
415. If the lowest power of x in the denominator is higher
than the lowest power of x in the numerator, the method of the
preceding article will he found inapplicable. We may, how
ever, determine by actual division what will be the exponent
of x in the first term of the expansion, and assume the fraction
equal to a series whose first term contains that power of x ; the
exponents afterwards increasing by unity each term as usual.
1. Expand ^ ; 2 into a series.
O X — X
Proceeding in the usual way, we should assume
1
=A+Bx+
Clearing of fractions, l = 3Ax + (3B — A)x 2 +
Equating the coefficients of like powers of x, we have 1 = ;
a result manifestly absurd, and showing that the usual method
is inapplicable.
x 1
But, dividing 1 by 3 x — x' 1 , we obtain — — as the first term
o
of the quotient ; hence we assume the fraction equal to a series
whose first term contains x~ l ; next term x°, or 1 ; next term
x ; etc. Or,
5 T = Ax~ 1 + B + Cx + Dx 2 + Ex*+
OX — X'
310 ALGEBRA.
Clearing of fractions, and collecting terms,
x* +
1=3A+3B
 A
x + 3C
 B
x + oD
 C
x 3 + 3E
 D
Equating the coefficients of like powers of x,
3 A = 1 ; whence, A =  .
o
3B~A = 0 whence, B = — =  .
3 9
7? 1
3 C—B — 0] whence, C = — = — .
3D C=0; whence, D = ^ = — .
3ED = 0; whence, E=^ = ~.
Substituting these values,
1 1.11 1.1,
3xx 2 3 9 27 '81 '243
EXAMPLES.
Expand the following to five terms :
2 n l + za 2 . l — 2x* — at
3sc 2 2a; 8 " cc2a 2 +3cc 3 a a + a: 8 a 4
APPLICATION TO THE EXPANSION OF RADICALS INTO SERIES.
416. As any root of any expression consisting of two or
more terms can be obtained by the method of Art. 247, we
know that the development is possible.
1. Expand \'l + x 2 into a series by the Theorem of Unde
termined Coefficients.
Assume \Jl\x' 2 = A + Bx + Cx*+ Dx* + 22x i +
UNDETERMINED COEFFICIENTS.
311
Squaring both members, we have (Art. 230),
l + x 2 = A 2
+ 2AB
x + B 2
+ 2 AC
.'■
+ 2 AD
+ 2BC
x 3 +
C 2
+ 2AE
+ 2BD
./■•
Equating the coefficients of like powers of x,
A 2 = l; whence, A = l.
2AB = 0; whence, B = 
B 2 + 2AC=1; whence, C =
2 A
1  B 2 _ 1
^2
2vlZ> + 2i?C Y = 0; whence, D = =0
C 2 + 2AB+2BD = 0; whence, .E=
2 A
BC
A
2BD+ C 2
2 A
1
8
Substituting these values,
yi + x 2 = l + x 2 ;x i +
2 8
which may be verified by the method of Art. 239.
Note. From the equation A = \, we may have A= ± 1 ; and taking
4 =  1, we should find C =  — , E=— , , so that the expansion might
2 8
he as follows :
2 8
This agrees with the remark made after the rule in Art. 239.
EXAMPLES.
Expand the following to five terms :
2. Sjl + x. 4. v / l2a; + 3x 2 .
3. v/12*. 5. y/1 + * + x 2 .
6. yiaj.
7. yl + ic + a; 2 .
[12 ALGEBRA.
APPLICATION TO THE DECOMPOSITION OF RATIONAL
FRACTIONS.
417. When the denominator of a fraction can be resolved
into factors, and the numerator is of a lower degree than the
denominator, the Theorem of Undetermined Coefficients ena
bles us to express the given fraction as the sum of two or more
'partial fractions, whose denominators are the factors of the
given denominator.
We shall consider only those cases in which the factors of
the denominator are all of the first degree.
CASE I.
418. When the factors of the denominator are all un
equal.
x + 7 .
Let — tt— 7z ^r be a fraction, whose denominator is
(3 x — 1) (5 x + 2)
composed of two unequal first degree factors. We wish to
prove that it can be decomposed into two fractions, whose
denominators are 3 x — 1 and 5 x + 2, and whose numerators
are independent of x. To prove this, assume
x + T A B
+
(3 x  1) (5 x + 2) 3 x  1 5 x + 2
We will now show that such values, independent of x, may
be given to A and B, as will make the above equation identi
cal, or true for all values of x. Clearing of fractions,
x + 7 = A (5 x + 2) + B (3 x  1)
or, x + 7 = (5 A + 3 B) x + 2 A  B,
which is to be true for all values of x. Then, by Art. 413, the
coefficients of like powers of x in the two members must be
equal ; or,
5^+35=1
2AB=7
UNDETERMINED COEFFICIENTS. 313
From these two equations we obtain A = 2, and B = — 3.
Hence, the proposed decomposition is possible, and we have
x + 7 2 3
(3xl)(5x + 2) 3xl'5x + 2
2
o
3a;l 5x + 2
This result may be readily verified by finding the sum of
the fractions.
In a similar manner we can prove that any fraction, whose
denominator is composed of unequal first degree factors, can
be decomposed into as many fractions as there are factors,
having these factors for their denominators, and for their nu
merators quantities independent of x.
EXAMPLES.
1. Decompose — 2 — j — — — into its partial fractions.
The factors of the denominator are x — 8 and x — 5 (Art. 118).
a .i 3 x — 5 A B ,* >.
Assume, then, — — — = \ (J)
x — Id x + 40 x — 8 x — 5
Clearing of fractions, and uniting terms,
3 x  5 = A (x  5) + B (x  8)
19
Putting x = 8, 19 = 3 A, or A = g
Putting cc = 5, 10= — 3 B, or B= — g
Note. The student may compare the above method of finding A and
B with that used on page 312.
Substituting these values in (1),
19 10
3^5 "3" 3_19 10
x 1  13 x + 40 ~~ xS + x — 5 ~ 3 (x  8) ~ 3 (x  5) ' AnS '
314 ALGEBRA.
EXAMPLES.
Decompose the following into their partial fractions
5ic — 2 3a; + 2 n cc
«•  o — r • ^ i — o — •
ar — 4 ar — 2 a;
x + 9 _ 2a; — 3
0.
6.
•>
ar —
13 a + 42 '
7.
17
6 a; 2
 13 x  5
a;
*
x 1 + 3 a; ' a; 2 — 3 a; — 4
ft 7a ' + 9 Q
9 + 9 x  4 a; 2 * * (a; 2  1) (x2)'
CASE II.
419. When the factors of the denominator are all equal.
x 2 — 11 x + 26
1. Separate ' 3 — into its partial fractions.
\x — o)
If we attempt to perform the example hy the method of
Case I, we should assume
x*—l±x + 2& ABC
+ o +
(x — 3) 3 x — 3 x — 3 x — 3
This would evidently he impossible, as the sum of the frac
tions in the second member is — ; which, as A, B,
x — o
and C are, by supposition, independent of x, cannot be equal
a; 2 11 a; + 26
tO ; jrr .
(x — 3) 8
The method to be used in Case II depends on the following :
n ., .,. a x" 1 + b x n ~ 2 + c x"* + + k
Consider the traction r .
(aj + h) n
Putting x = y — h, the fraction becomes
a(y hy 1 + b(i/ — h) n  2 + e (y — A)"" 3 + + k
UNDETERMINED COEFFICIENTS. 315
If the terms of the numerator are expanded by the binomial
theorem, and the terms containing like powers of y collected
together, we shall have a fraction of the form
a, y' 1 ' 1 + b r y n ~ 2 + c x y n ~ 3 + +ky
y ,
Dividing each term of the numerator by y n , we have
a x b x c x l\
y y y v
Changing back y to x + h, this becomes
«i b, c, /.',
+ , , ' , + , ,' Q + +
x + h (x+ h) 2 (x + h) 3 (x + h) n
This shows that the assumed fraction can be expressed as
the sum of n partial fractions, whose numerators are indepen
dent of x, and whose denominators are the powers of x + h,
beginning with the first, and ending with the nth.
In accordance with this, we assume
a 2 11 a; + 26 A B , C
( x 3) 3 ~ x  3 T (x  3) 2 ^ (x  3) 3 '
Clearing of fractions,
x 2 llx + 26 = A(x3) + B(x3)+ C
= A (x 2 6x + 9) + B(x — B)+C
= A x 2 + (B  6 A) x + 9 A  3 B + C.
Equating the coefficients of like powers of x,
A = l, B6A = 11, and 9A3B+ C=26
Whence, A = 1, B = 5, and C = 2.
Substituting these values,
a; 2 lla; + 26_ 1 5 2
(x  3) 3 ~~ " ~x~^3 ~ (x  3) 2 + (x  3) 3 ' US '
2'
316 ALGEBEA.
EXAMPLES,
Separate the following into their partial fractions :
ar+3a: + 3 a; 2 3 a; — 10
(x + 1) 3 " ' (x2) 3 * (2a5) s
2 a; 13 3 a; 2 4 18 a; 2 + 12 a;  3
(x5) 2 ' (cc + 1) 8 ' (3 x + 2/
CASE III.
420. When some of the factors of the denominator are
equal.
1. Separate zrrz into its partial fractions.
1 x (x + l) 3 l
The method in this case is a combination of the methods of
Cases I and II. We assume
3 a + 2 A B C D
+ ,.. . ,,, + ,.. , ^„ + — •
a; (x + l) 3 a + 1 (x + l) 2 (a; + l) 3
Clearing of fractions,
3ai + 2 = A x (x + l) 2 + B x {x + 1) + Cx + D (x + l) 3
= (A + D) x* + (2A + B + 3 D) x 2 + ( A +B + C+ 3 D) x +B.
Equating the coefficients of like powers of x,
£> = 2, A + B + C+3 B = 3, 2A + B+3B = 0, and
A + B =
Whence, A = 2, B = 2, C = l, andZ> = 2.
Substituting these values,
3a; + 2 2 2 12,
a;(a; + l) 3 a; + 1 (a? + 1) 2 (a + l) 3 x
It is impossible to give an example to illustrate every pos
sible case; but no difficulty will be found in assuming the
UNDETERMINED COEFFICIENTS. 317
proper partial fractions, if attention be given to the following
general case. A fraction of the form
X
(x + a) (x + b) (x + m) r (x + n) s
should he put equal to
A B E t F K
x+a x+b x + m (x+m)' (x + m) r
L M R
+ x + n + (x + n) 2 + + (x + n) s +
Single factors, like x + a and x + b, having single fractions
A B
like and  — — , corresponding ; and repeated factors,
x+a x+b
like (x + m) r , having r partial fractions corresponding, ar
ranged as in Case II.
EXAMPLES.
Separate the following into their partial fractions :
83 xx' 2 15  7 x + 3 x 2  3 x z
x(x + 2) 2 ' x 3 (x + 5)
3 ,r 3  11 x 2 + 13 x  4 6 ar  14 x + 6
x (x  1) (x  2) 2 ' (x2)(2x3y 2 '
3a;l ? 5 x 2 + 3 x + 2
x 2 {x + iy ' x 3 (x + iy 2 '
421. Unless the numerator is of a lower degree than the
denominator, the preceding methods are inapplicable.
:' .'• + 1 .
For example, let it be required to separate — ^— — into its
partial fractions. Proceeding in the usual way, we assume
2 .r +1 A B
Ob *jC JC *fc x.
Clearing of fractions,
2 x 2 + 1 = A (x  1) + B x = (A + B) x  A.
318
ALGEBRA.
Equating the coefficients of like powers of x, we have 2 = 0;
an absurd result, and showing that the usual method is inap
plicable.
But by actual division, as in Art. 150, we have
2z 2 +l 2« + l
— £ +
x — x
x
X
2 x + 1 . .,„.,
We may now separate — ^ into its partial fractions by
the usual method, obtaining
2a; + 1 1 3
x' — x
X X
1"
Hence,
2a 2 +l 2^ + 1 1 3
= 2 + —r =2  +  —,Ans.
x
X
x — X
APPLICATION TO THE REVERSION OF SERIES.
422. 1. Given y = 2x + x 2 — 2x z — 3x* + , to revert
the series, or to express x in terms of y.
Assume x = Ay + B if + G if + D if + (1)
Substituting in this the given value of y, we have
x=A(2x + x 2 2x s 3x i +...)+B(ix 2 + x 4 +Ax s Sx i + ...)
+ <7(8x 3 + 12z 4 +...) + Z>(16z 4 +...) +
or,
X:
2Ax+ A
x 2 2A
x 3  3 A
+ 4:B
+ 4B
 IB
+ 8
+ 12 G
+ 16Z>
x* +
Equating the coefficients of like powers of x,
2 A = 1 ; whence, A =  .
A + 4 B— ; whence, B— r = —  ,
4 8
2A + 4.B+S G=0; whence, 0=
3^
16"
3^7^+12 C+1GZ> = 0; whence, D —
13
128'
UNDETERMINED COEFFICIENTS.
319
Substituting these values in (1),
V if 3 y 3 13 y i
If the even powers of x are wanting in the given series, we
may abridge the operation by assuming x equal to a series
containing only the odd powers of y.
Thus, to revert the series y = x — x 3 + x 5 — x 1 + , we
assume x = A y + B y 3 + C y 5 + D y" +
If the odd powers of x are wanting in the given series, the
reversion of the series is impossible by the method previously
given. But by substituting another letter, say t, for x~, we
may revert the series and obtain a value of t, or of x 2 , in terms
of y ; and by taking the square root of the result, express x
itself in terms of y.
If the first term of the series is independent of x, we cannot,
by the method previously given, express x definitely in terms
of y ; though we can express it in the form of a series in which
y is the only unknown quantity.
2. Kevert the series y = 2 + 2x — x 2 — x 3 + 2x i +
We may write the series,
y2 = 2xx 2 x 3 + 2x 4 + (1)
Assume x=A(y2) + B(y2y 2 +C(y2) 3 +D(y2y+... (2)
Substituting in this the value of y — 2 given in (1), we have
x = A(2x— x 2 — x s +2x*+ ...) + B(4:X 2 + x i — ix 3 — 4x*+ ...)
+ C(Sx 3 12x i + ...) + D(lGx 4 + ...)+
or, x = 2 A x — A
+ 4:B
x 2  A
x 3 + 2 A
42?
 3B
+ SC
12C
+ 16B
x* +
320 ALGEBRA.
Equating the coefficients of like powers of x,
2 A — l; whence, A = ^ .
— A + 4 B =:Jd ; whence, B = ^ .
o
— A4:B + 8C = 0; whence, C = ^.
o
7
2A3B12C+16I> = 0] whence, D = j.
Suhstituting in (2),
x=l(y2)+l( 1/ 2y+l( }/ 2y+^( u 2y+ ,Am.
EXAMPLES.
Revert the following series to four terms :
3. y = x + x + x 3 + x i +
4. ij = 2x + 3x 3 + 4:x 5 + 5x 7 +
5. i/ = x — x 3 + x 5 — x 1 +
2 3 4
^y»— rj/*" 'V*^
\K/ *Aj *AJ
tAs \Kj \Aj
8. y = 3x2x 2 +3x 3 4x i +
Note. This method may sometimes be used to find, approximately, the
root of an equation of higher degree than the second. Thus, to solve the
equation
we may put .1 = 2/, and revert the series ; giving, as in Ex. 1, Art. 422,
1 1 „ 3 , 13
x = — ii if\ if if +
2 & J 16 128
Putting back y=.l, we have
.1 .01 .003 .0013
^ 2 8 16 128
= . 05. 00125 + . 00019 . 00001 + = .04893 + , Ans.
This method can, of course, only be used when the series in the second
member is convergent.
BINOMIAL THEOREM. 321
XXXIX. — BINOMIAL THEOREM.
ANY EXPONENT.
423. We have seen (Art. 402) that when n is a positive
integer,
n(n— 1) „ n(n — l)(n — 2) „
(l + a; )»=l + wa; + — ^r V+ — ^ ' x z +
We shall now prove that this formula is true when n is a
positive fraction, a negative integer, or a negative fraction.
1. Let n be a positive fraction, which we will denote by
p ...
— ; p and q being positive integers.
Now (Art. 252), (1 + x) * = V (1 + xf
= yi+px+ , (Art. 402).
Extracting the ^th root of this expression by the method of
Art. 247,
1 +px +
1« = 1
. p x
1 + — +
q j) x
That is, (l + a:)i" = l + — + (1)
2. Let n be a negative quantity, either integer or fraction,
which we will denote by — s.
Then (Art. 255),
= , (by Arts. 402, and 423, 1).
1 + sx+ ' v J
From which, by actual division, we have
(l + x)° = lsx + (2)
322 ALGEBRA.
From (1), (2), and Art. 402, we observe that whether n is
positive or negative, integral or fractional, the form of the
expansion is
(1 + x) n = 1 + n x + A x 2 + B x 3 + Cx* + (3)
x
Writing  in place of x, we have
a
/y* /y>J /y.3 /y»4
1 + ) =l + n + A~ r2 + B+C— i +
al a a 1 a 6 a 4
Multiplying this through by a n , and remembering that
(x\ n r / x\ i "
1+) = a\l+j = (a + x) n , we have
(a + x) n = a n + n a"' 1 x + A a n ~ 2 x + Ba n ~ s x s + (4)
To find the values of A, B, etc., we put x + z for x in (3),
and regarding (x + z) as one term, we shall have
[1 + (x + z)'] r ' = l + ii(x + z) + A(x + z) 2 + B(x + z) 3 +
= 1 + n x + A x' 1 + B x s +
+ (n + 2Ax + 3Bx 2 + )*+ (5)
Regarding (1 + x) as one term, we shall have, by (4),
[(1 + x) +z~] n = (1 + x) n + n (l + x) n  1 z+ (6)
Since [1 + (x + z)~\ n = [(1 + x) + z\ n , identically, we have
from (5) and (6),
l + nx + Ax 2 +Bx z + + (n + 2Ax + 3Bx 2 + )z+
= (1 + a) n + w(l + a) n_1 * +
which is true for all values of z which make both members of
the equation convergent. Hence, by Art. 413, the coefficients
of z in the two series must be equal ; or,
11 (l + x) n  1 = n + 2Ax + 3Bx 2 +
BINOMIAL THEOREM. 323
Multiplying both members by 1 + x,
n(l + x) n = n+ (2A + n)x+(3B+2A)x 2 +
or, by (3),
n + n 2 x + n Ax 2 + n Bx 3 + = n+ (2 A + n) x
+ (3B+2A)x 2 +
•which is true for all values of x which make both members of
the equation convergent ; hence, equating the coefficients of
like powers of x,
sfy {fl, j\
2 A + n = n 2 ; whence, 2 A = n 2 — n, or A =  =
3B+2A = ?iA; whence, 3 B=n A2 A = A (n2)
„ A (n  2) n (n  1) (n  2)
B =~3— = \3
Substituting in (4),
(a + x) n = a n + na" 1 x+ V ^\~ ' a n ~ 2 x 2
n(nl)(n2)
+ — i ^ a n ~ a X s +
which has tbus been proved to hold for all values of n, positive
or negative, integral or fractional. Hence, the Binomial The
orem has been proved in its most general form. The result,
however, only expresses the value of {a + x) n for such values
of x as make the second member convergent (Art. 413).
424. When n is a positive integer, the number of terms in
the expansion is n + 1 (Art. 399). When n is a fraction or
negative quantity, the expansion never terminates, as no one
of the quantities n — 1, n — 2, etc., can become equal to zero.
The development in that case furnishes an infinite series.
324 ALGEBRA.
425. The method and notes of Art. 403 apply to the ex
pansion of expressions hy the Binomial Theorem when the
exponent is a fractional or negative quantity.
2
1. Expand (a + x) :! to five terms.
2
The exponent of a in the first term of the expansion is  , and
o
decreases hy one in each succeeding term.
The exponent of x in the second term of the expansion is 1,
and increases hy one in each succeeding term.
The coefficient of the first term is 1 ; of the second term,
2 2
 ; multiplying the coefficient of the second term, , hy the ex
o o
1 2
ponent of a m that term, —  , and dividing the product, —  ,
hy the numher of the term, 2, we ohtain —  as the coefficient of
the third term ; etc.
2 2_i 1 _4 4 i 7 _jo
Kesult, a 3 + ~a 3 x—a ^x 2 +^ra 3 x 3 — 7 rr^a s x A \
.^12
2. Expand (1 + 2 a: 2 ) 2 to five terms.
(l + 2zV 2 =D + (2*)]
= l 2  2.1= . (2 x$) + 3.1 4 . (2 xfy  4.1 5 . (2 a;*) 8
+ 5.1 6 . (2 a;*) 4
i
= 1  2 (2 x) + 3 (4 a)  4 (8 x?) + 5 (16 a 2 )  ....
= l4a 2 + 12x32a: + S0a; 2 — , Ans.
3. Expand (a 1 — 3 a: *) * to five terms.
(a 1 3a 5 ~ i )~* = [(«" 1 ) + (3 a;  *)]"*
BINOMIAL THEOREM. 325
= (a 1 )"^  \ (co 1 )'" ( 3 aT*) + " (0~ V ( 3 x~fy
1 40 L3 _ JL 455 —JUS 1
TFr(0~ v (3* l ) 8 +iS(« 1 ) ¥ (3* *)«
81 v y v y ' 243
447 _i 14 lo 140 13 _a
= ft 3_* :w 3a; 2 ) + 3 (9 a; i)__^ a V(_27 a; 2)
O 9 ol
+i*>o
4 7 _i J" 140 13 _a 455 .lb
= a^+4a 3 a; 2 + 14a 3 cc _1 + — — a 3 x 2 + —  a 3 #~ 2 ,
o o
^l?zs.
EXAMPLES.
Expand the following to live terms :
4. O + z) 2 8. . 8/r — — ' 12. (m~"3_2« 2 ) 2 .
v y yl + a; v '
5. (1 + ^) 6 . 9. ^^y 3  13. (l + e.r 1 )^.
6. (lx)~%. 10. — — . 14. (a; 4 + 4aJ)t
c 2 + d
7. V^ 1 ^ 11. (aT*8y)*. 15 (g i_ 8y y
426. The expression for the rth term, derived in Art. 404,
holds for any value of n, as it was deduced from the expansion
which has been proved to hold universally.
1. Find the 7th term of (1 — cc)~ 3 .
Here r = 7, n = — ^ ; hence, the
o
1 4 _7 10 13 16
~3 ,_ 3*~3'""3"' ' "3"" " 3 . NR 728a; 6
j th term = 17 2.3.4.5.6 ( ~ x) = "656T '
Ans.
326 ALGEBRA.
2. Find the 8th term of (eft + x~%)~ z .
Here r = 8, n = — 3 ; hence, the
3.4.5.6. — 7.8.9 , ^ „ . _* ,
8thterm = 0.3.4.5.6.7 ^ W (* ^
= — 36 ar b x 3 , Ans.
EXAMPLES.
Find the
3. 8th term of \/ a + x. 7. 7th term of (x' 1 — ifift.
4. 7th term of (1 + m) 4 . 8. 5th term of .
(n~* — <ry
5. 5th term of (1 — ar)~ *. 9. 6th term of (eft + 3 x~ l )~%.
1 2
6. 6th term of . 10. 8th term of (x 3 y — z 3 ) 3 .
y'x 2 + f
427. To find any root of a number approximately by the
Binomial Theorem.
1. Find the approximate square root of 10.
y/ 10 = 10* = (9 + 1) * = (3 2 + 1)*
Expanding this hy the Binomial Theorem,
(3' + 1)* = (3*)* + \ (3 a )i  \ (3')* + 1 (3T*
111 ^
— 3 _L Z 31 _ Z Q8 i Q5 ^ q7 i
6 + 2 .6 8 .3 +jg. 3 J283 +
= 3i _1_ J_ _t_ _JL
+ 2.3 8.3 3 + 16.3 6 128. 3 7+
= 3 + .16667  .00163 + .00026  .00002 +
= 3.16228+,
BINOMIAL THEOREM. 327
which is the approximate square root of 10 to the fifth decimal
place, as may be verified by evolution.
2. Find the approximate cube root of 26.
$ 26 = 26* = (27 1)3 = (3 3  1)*
Expanding this by the Binomial Theorem,
(3 3  1)* = (3 3 )^ + \ (3 3 ) 3 * ( 1)  i (3 3 )"* ( 1)»
+ [ (3 3 r l (i) 3 
1 1 K
Q Q2 Q5 Q8
°~r 6 ~9 "si" 5 ~
= 3
3.3 2 9.3 5 81. 3 8
= 3  .037037  .000457  .000009 
= 2.962497 + , Ans.
RULE.
Separate the given number into two parts, the first of whir//
is the nearest perfect power of the same degree as the required
root. Expand the result by the Binomial Theorem.
Note. If the second term of the binomial is small, the terms in the
expansion converge rapidly, and we obtain an approximate value of the
required root by taking the sum of a few terms of the development. But
if the second term is large, the terms converge slowly, and it requires the
sum of many terms to insure a considerable degree of accuracy.
EXAMPLES.
Find the approximate values of the following to five deci
mal places :
3. #31. 5. #99. 7. #17.
4. #9. 6. #29. 8. #78.
328 ALGEBRA.
XL. — SUMMATION OF INFINITE SERIES.
428. The Summation of a Series is the process of finding
a finite expression equivalent to the series.
Different series require different methods of summation,
according to the nature of the series, or the law of its forma
tion. Methods of summing arithmetical and geometrical series
have already been given (Arts. 369, 377, and 380). Methods
applicable to other series will now be treated.
RECURRING SERIES.
429. A Recurring Series is one in which each term, after
some fixed term, bears a uniform relation to a fixed number of
the preceding terms. Thus
l + 2x + 3x 2 + Ax 3 +
is a recurring series, in which each term, after the second, is
equal to the product of the preceding term by 2 x, plus the
product of the next term but one preceding by — x 2 .
The sum of these constant multipliers is called the scale of
relation of the series, and their coefficients constitute the scale
of relation of the coefficients of the series. For example, in
the series 1 + 2 x + 3 x' 2 + 4 x s + , the scale of relation is
2 x — x 2 , and the scale of relation of the coefficients is 2 — 1.
430. A recurring series is said to be of the first order
when each term, commencing with the second, depends on the
one immediately preceding; of the second order, when each
term, commencing with the third, depends ujion the tiro im
mediately preceding ; and so on.
If the series is of the first order, the scale of relation will
consist of one term ; if of the second order, it will consist of
two terms ; and, in general, the order and the number of terms
in the scale of relation will correspond.
431. To find the scale of relation of the coefficients of a
recurring series.
SUMMATION OF INFINITE SERIES. 329
1. If the series is of the first order, it is a simple geometri
cal progression, and the scale of relation of the coefficients is
found by dividing the coefficient of any term by the coefficient
of the preceding term.
2. If the series is of the second order, let a, b, c, d,
represent the consecutive coefficients of the series, and p + q
their scale of relation. Then,
c =p b + q a)
d=p c + q b j
(4>
to determine p and q ; solving, we obtain
ad — bc .. c 2 — b d
p = 77 , and q = — — .
ac — b 1 ac — b 1
3. If the series is of the third order, let a, b, c, d, e, f,
represent the consecutive coefficients of the series, and p + q
+ r their scale of relation. Then,
d =p c + q b + r a
e =p d + q c + r b
f=X> e + q d + r c
from which we can find p, q, and ;■.
432. To ascertain the order of a series, we may first make
trial of a scale of two terms, and if the result does not corre
spond with the series, we may try three terms, four terms, and
so on, till the true scale of relation is found. If we assume
the series to be of too high an order, the terms of the scale
will take the form r. •
433. To find the sum of a recurring series, when the scale
of relation of its coefficients is known.
Let
a + bx + cx 2 + dx 3 + +jx n ~ 3 + kx n ~ 2 + lx n ~ l +
be a recurring se"ries of the second order. Let S denote the
330 ALGEBKA.
sum of n terms of the series ; and let p + q be the scale of re
lation of the coefficients. Then,
S= a + b x + c x 2 + d x 3 + + lx n ~ x
p Sx=pax+pbx 2 +pcx 3 + + pkx n ~ 1 +plx n
q Sx 2 = q ax 2 + q b X s + + qjx n ~ 1 +q k x n + qlx n + 1
Subtracting the last two equations from the first,
S—p Sx — q Sx 2 = a + bx — pax— plx n — q kx n — qlx n + 1
the rest of the terms of the second member disappearing, be
cause, since p + q is the scale of relation of the coefficients,
c =p b + q a, d =p c + q b, I =p k + qj.
Therefore we have
a + (b—p a) x — (p I + q k) x n — q I x n + 1
S =
■px — q x"
the formula for finding the sum of n terms of a recurring series
of the second order.
But if n becomes indefinitely great, and the series is con
vergent, then the limiting values of the terms which involve
x n and x n + ! must become 0, and we have at the limit
s= a + (ppa)x
1 — p x — qx 2
the formula for finding the sum of an infinite recurring series
of the second order.
If q = 0, then the series is of the first order, and conse*
quently b =p a; then,
1 —px
the formula for finding the stun of an infinite recurring series
of the first order. (Compare Art. 380.)
SUMMATION OF INFINITE SERIES. 331
In like manner, we should obtain
a + (b — p a) x + (c —pb — qa)x 2
o — z o 5 (o)
1 —p x — q x — r x 6 w
the formula for the summation of an infinite recurring series
of the third order.
434. A recurring series, like other infinite series, originates
from an irreducible fraction, called the generating fraction.
The summation of the series, therefore, reproduces the frac
tion ; the operation being, in fact, the exact reverse of that in
Art. 414.
435. 1. Find the sum of l + 2cc + 8ar+28z 3 +100x 4 +
We must first determine the scale of relation of the coeffi
cients. In accordance with Art. 432, we first assume the
series to be of the second order. We have a = i, b = 2, c==8,
d = 28. Substituting in the values of p and q derived from
(^4), Art. 431, we have p = 3 and q = 2. To ascertain if this
is the proper scale of relation, consider the fifth term, 100 x 4 ;
this should be 3 x times the preceding term, plus 2 x 2 times
the next preceding term but one, or, 84 x 4 + 16 x*. This shows
that the series is of the second order.
Substituting in (1) the values of a, b, p, and q, we have
l + (2 — 3)a; _ 1 — x
b ~ iZx2tf ~ l3x2x 2 '
EXAMPLES.
Find the sum of the following series :
2. l + 2a + 3x 2 + 5a; 3 + 8a; 4 +
a ac a c 2 „ a c 3
3> b^¥ x + !F x "l^ x+
4. 4 + 9a; + 21ar+51;r 5 +
5. l + 3x + 5x 2 + 7x 3 +
332 ALGEBRA.
6. 2a + 2a 2 5a 3 +10a i 17a 5 +
7. 3 + 5x + 7 x' 2 + 13.x' 3 + 23 a; 4 + 45 x* +
8. l + 3x + 4:x 2 +7x 3 +llx 4 +
9. 2 + 4xx*3x s + 2x i + ±x 5 +
DIFFERENTIAL METHOD.
436. The Differential Method is the process of finding any
term, or the sum of any number of terms, of a regular series,
by means of the successive differences of its terms.
437. If, in any series, we take the first term from the sec,
ond, the second from the third, the third from the fourth, and
so on, the remainders will form a new series called the first
order of differences.
If the differences be taken in this new series in like manner,
we obtain a series called the second order of differences ; and
so on.
Thus, if the given series is
1, 8, 27, 64, 125, 216,
the successive orders of differences will be as follows :
1st order, 7, 19, 37, 61, 91,
2d order, 12, 18, 24, 30,
3d order, 6, 6, 6,
4th order, 0, 0,
Hence, in this case there are only three orders of differences.
438. To find any term of a series.
Let the series be
a l) a 2> a 3> a i) a 5> a n1 a n+l>
Then the first order of differences will be
SUMMATION OF INFINITE SERIES. 333
the second order of differences will be
a 3 — 2a 2 + a l , a A — 2a 3 + a 2 , a 5 — 2a i + a 3 , ,
the third order of differences will be
a A — 3 a 3 + 3 a 2 — a x , a 5 — 3 a A + 3 a 3 — a 2 , ,
the fourth order of differences will be
a 5 — 4 a A + 6 a& — 4 a 2 + a 1} ,
and so on ; where each difference, although a compound quan
tity, is called a term.
Let now d x , d 2 , d s , d A , represent the first terms of the
several orders of differences. Then,
d l = a 2 — a l ; whence, a 2 = a l + d A .
d 2 = a 3 — 2 a 2 + a x ; whence, a 3 — 2a 2 — a l +d 2 ^2ai + 2d 1
— a x + d 2 = a A + 2 d A + d 2 .
d 3 = a A — 3 a s + 3 a 2 — a x ; whence, a A = a x + 3 d v + 3 d 2 f d s .
d A = a 5 — A a A + 6 a 3 — Aa 2 +a l ; whence, a 5 = a l + 4^d 1 + 6d 2
+ ±d 3 +d A .
We observe that the coefficients of the value of a 2 are the
same as the coefficients of the first power of a binomial; the
coefficients of the value of a 3 are the same as the coefficients
of the second power of a binomial ; and so on. Assume that
this law holds for the nth term ; that is, that the coefficients
of the value of a H are the same as the coefficients of the (n — l)th
power of a binomial ; then,
, „ . (nl)(n2) .
a* = «i + (n  1) (h + Q d 2
+ („!)(„ 2) <«S) d3+ (1)
If the law holds for the nth term in the given series, it will
also hold for the «th term in the first order of differences ; or,
334 ALGEBRA.
a n+1 a n = d 1 +(nl)d 2 + ^—^f—^d ;i + (2)
Adding (1) and (2), we have
(n  1) (n  2)
a
, +1 = «!+[! + (nl)]d l +
(n  1) +
+
(wl)(w2) (wl)( w 2)(ra3)
~2 ~\T
2
,/.,
= aj + n dx H r^ [2 + n — 2] d 2
(n  1) («  2) ro
, n(nl) ? »(»l)(n2) . ,„
= «! + »!(/!+ ^ ~ <4 + ^ "^3 + (3)
where the coefficients are the same as the coefficients of the rath
power of a binomial. Hence, if the law holds for the nth term,
it also holds for the {n + l)th term ; but we have shown it to
hold for the fifth term, a 5 ; hence it holds for the sixth term ;
and so on. That is, Formula (1) holds for any term in the
series.
When the differences finally become 0, the value of the nth
term can be obtained exactly ; but, in other cases, the result is
merely an approximate value.
439. To find the stem of any number of terms of a series.
Let the series be
a, b, c, d, e, (1)
Let S denote the sum of the first n terms. Assume the
series
0, a, a + b, a + b + c, a + b + c + d, (2)
in which the (n + l)th term is obviously equal to the sum of n
terms of the given scries ; that is, S is the (n + l)th term of
series (2). Now the first order of differences of series (2) is
SUMMATION OF INFINITE SERIES. 335
the same as series (1) ; hence, the second order of differences
of series (2) is the same as the first order of (1) ; the third
order of (2) is the same as the second order of (1) ; and so on.
Then, letting a', d\, d' 2 , d' s , represent the first term, and
the first terms of the several orders of differences of (2), we
have a' = 0, d\ = a, d' 2 = d l ,d' s = d 2 , where a, d x , d 2 ,
are the first term, and the first terms of the several orders of
differences of (1). But, by (3), Art. 438, the (n + l)th term
of series (2) will be
n (n  1) „ n (n  1) (n  2)
In this put for a', d\, d' 2 , d' 3 , their values; then
n (n — 1) , n (n — 1) (re — 2) .
S=na + — ^2 — L d 1 + ~" L d,+ (3)
440. 1. Find the 12th term of the series 2, 6, 12, 20,
30,
The successive orders of differences will be as follows :
1st order, 4, 6, 8, 10,
2d order, 2, 2, 2,
3d order, 0, 0,
Then a x = 2, d Y = 4, d 2 = 2, d s , d A , = 0, and n = 12.
Substituting in (1), Art. 438, the 12th term
(12 — 1) (12 — 2)
= 2 + (12l)4+ l >) } ; 2 = 2 + 44 + 110=156,^.
2. Find the sum of 8 terms of the series 2, 5, 10, 17,
1st order of differences, 3, 5, 7,
2d order of differences, 2, 2,
3d order of differences, 0,
Then a = 2, d, = 3, d, = 2, n = S.
336 ALGEBRA.
Substituting these values in (3), Art. 439, we have
= 16 + 84 + 112 = 212, Ans.
EXAMPLES.
3. Find the first term of the fifth order of differences of
the series 6, 9, 17, 35, 63, 99,
4. Find the first term of the sixth order of differences of
the series 3, 6, 11, 17, 24, 36, 50, 72,
5. Find the seventh term of the series 3, 5, 8, 12, 17,
6. Sum the first twelve terms of the series 1, 4, 10, 20,
35,
7. Sum the first hundred terms of the series 1, 2, 3, 4,
5,
8. Find the 15th term of the series l 2 , 2 2 , 3 2 , 4 2 ,
9. Sum the first n terms of the series l 3 , 2 3 , 3 3 , 4 3 , 5 3 ,
10. Sum the first n terms of the series 1, 2 4 , 3 4 , 4 4 , 5 4 , 6 4 ,
11. If shot be piled in the shape of a pyramid, with a trian
gular base, each side of which exhibits 9 shot, find the number
contained in the pile.
12. If shot be piled in the shape of a pyramid, with a square
base, each side of which exhibits 25 shot, find the number
contained in the pile.
INTERPOLATION.
441. Interpolation is the process of introducing between
terms of a series other terms conforming to the law of the
series.
SUMMATION OF INFINITE SERIES. 337
Its usual application is in finding intermediate numbers
between tbose given in Mathematical Tables, which may be
regarded as a series of equidistant terms.
442. The interpolation of any intermediate term in a
series, is essentially finding the nth term of the series, by the
differential method (Art. 438). Thus,
Let t represent tbe term to be interpolated in a series of
equidistant terms, and p the distance the term t is removed
from the first term, a, expressed in intervals and fractions of
an interval ; that is, p being the distance to the nth term,
p = n — 1 intervals.
In Formula (1), Art. 438, putting p for n — 1, the nth. term
t=a+pdl+ pj£ ! =v. il+ p(* 1 np*> dt+
443. 1. In the series :j^ , 77 > TE > T£ > T?? > , find the
13 14 15 16 17
middle term between 5= and ^7 .
15 lb
Here, the first differences of the denominators are
1, 1, 1, 1,
The second differences are
0, 0, 0,
Whence, d t = 1, and d 2 = 0.
5
The distance to the required term is 2\ intervals, or p = ^
Make a = 13, the denominator of the first term ; then by the
preceding formula, the denominator of the required term,
1 2
Therefore the required term is — or 757, Ans.
31 ol
~2
338 ALGEBRA.
2. Given ^94 = 9.69536, ^95 = 9.74679, y/ 96 = 9.79796 ;
to find y/94i.
Here, the first differences are
.05143, .05117,
and the second differences are
.00026,
Whence, ^ = .05143, d 2 = . 00026,
. 1 . 1
The distance of the required term is  interval, or p = ? ,
Then the required term,
1(1
* = 9.69536 + 1 x .05143 + 4 ^ ( ,0(
= 9.69536 + .01286  J, ( .00026) +
= 9.69536 + .01286 + .00002 +
= (approximately) 9.70824, Ans.
EXAMPLES.
3. Given ^64 = 4, ^65 = 4.0207, f 66 = 4.0412, ^67 =
4.0615 ; find ^66A
4. Given ^45 = 3.556893, ^47 = 3.608826, ^49 = 3.659306,
^51 = 3.708430; find ^48.
5. Given ^5 = 2.23607, yf 6 = 2.44949, ^7 = 2.64575, y/8
= 2.82843 ; find \'r>M.
6. (Jivcn the length of a degree of longitude in latitude
41°=4528 miles; in latitude 42° = 44.59 miles; in latitude
43°=43.8S miles; in latitude 44° = 43.16 miles. Find the
length of a degree of longitude in latitude 11' 30'.
LOGARITHMS. 339
7. If the amount of $ 1 at 7 per cent compound interest for
2 years is $ 1.145, for 3 years $ 1.225, for 4 years $ 1.311, and
for 5 years $ 1.403, what is the amount for 4 years and 6
months ?
XLI. — LOGARITHMS.
444. The logarithm of a quantity to any given base, is the
exponent of the power to which the base must be raised to equal
the quantity.
For example, if a x = m, x is the exponent of the power to
which the hase, a, must be raised to equal the quantity, m;
or, x is the logarithm of m to the base a ; which is briefly
expressed thus :
x = log a m.
445. If a remain fixed, and m receive different values, a
certain value of x Avill correspond to each value of m; and
these values of x taken together constitute a System of Loga
rithms. And as the base, «., may have any value whatever, the
number of possible systems is unlimited.
For example, suppose a = 3.
Then, since 3° = 1, by Art. 444, = log., 1
" 3 X = 3, " " l = log 3 3
" 3 2 =9, " " 2 = log 3 9
Hence, in the system whose base is 3, log 1 = 0, log 3 = 1,
log 9 = 2, etc.
Again, suppose a = 12.
Then, since 12 1 = 12, 1 = log 12 12
" 12 2 = 144, 2 = log 12 144
Hence, in the system whose base is 12, log 12 = 1, log
144 = 2, etc.
340 ALGEBEA.
446. The only system in extensive use for numerical com
putations is the Common System or Briggs' System, whose
base is 10. Therefore the definition of the common logarithm
of a quantity is the exponent of that power of 1.0 which equals
the quantity. Hence,
Since 10° = 1, log 10 1 =
10 1 = 10, log 10 10 = 1
" 10 2 =100, log 10 100 = 2
TO 3 = 1000, log 10 1000 = 3
" 10 1 = i = .l, log in .l = l
a
10 2 = ^ = .01, log 10 .01=2
'< 10 3 ^ = .001, log 10 .001 = 3, etc.
447. It is customary in using common logarithms to omit
the subscript 10 which denotes the base ; hence, we may write
the results of Art. 446 as follows :
log 1 = log .1= 1 = 9 10
log 10 = 1 log .01 =  2 = 8  10
log 100 = 2 log .001 = 3 = 7 10
log 1000 = 3 etc.
The second form of the results in the second column will be
found less complicated in the solution of examples.
448. We infer the following from the first column of
Art. 447 :
The logarithm of any number between 1 and 10, lies between
and 1.
The logarithm of any number between 10 and 100, lies be
tween 1 and 2.
LOGARITHMS. 341
The logarithm of any number between 100 and 1000, lies be
tween 2 and 3, etc.
Or, in other words,
The logarithm of any number with one figure to the left of
its decimal point, is equal to plus some decimal.
The logarithm of any number with two figures to the left of
its decimal point, is equal to 1 plus some decimal.
The logarithm of any number with three figures to the left
of its decimal point, is equal to 2 plus some decimal, etc.
449. Reasoning in the same way from the second column
of Art. 447,
The logarithm of any number between 1 and .1, lies between
and 9 — 10, or between 10 — 10 and 9 — 10.
The logarithm of any number between .1 and .01, lies be
tween 9 — 10 and 8 — 10.
The logarithm of any number between .01 and .001, lies be
tween 8 — 10 and 7 — 10, etc.
Or, in other words,
The logarithm of any decimal with no zeros between its
point and first figure, is equal to 9 plus some decimal — 10.
The logarithm of any decimal with one zero between its
point and first figure, is equal to 8 plus some decimal — 10.
The logarithm of any decimal with two zeros between its
point and first figure, is equal to 7 plus some decimal — 10,
etc.
450. It will be seen from the two preceding articles that
in general the logarithm of a number consists of two parts,
one integral, the other decimal. The integral part is called
the characteristic ; the decimal part, the mantissa. For rea
sons which will be given hereafter, only the mantissa of the
logarithm is given in the tables ; the characteristic must be
supplied by the reader. The rules for characteristic are based
on the results obtained in the last parts of Arts. 448 and 449.
342 ALGEBRA.
451. I. If the number is greater than 1, the characteristic
is 1 less than the number of figures to the left of the decimal
[lit'; lit.
For example, characteristic of log 354.89 = 2,
characteristic of log 906328.3 = 5, etc.
II. If fin' number is less than 1, the characteristic is found
by subtracting the number of zeros between the decimal point
cut/ first significant figure from 9; writing — 10 after the
mantissa.
For example, characteristic of log .00792 = 7, with — 10
after the mantissa; characteristic of log .2583 = 9, with —10
after the mantissa ; etc.
It is customary in ordinary computation to omit the — 10
after the mantissa ; it should he remembered, however, that it
is really a part of the logarithm, and should be allowed for,
and subjected to precisely the same operations as the rest of
the logarithm. Beginners will find it useful to write it in
all cases ; and in some problems it cannot conveniently be
omitted.
Note. Many writers, in dealing with the characteristics of the loga
rithms of numbers less than 1, combine the two portions of the characteris
tic, writing the result as a negative characteristic before the mantissa.
Thus, instead of such an expression as 7.60358210, the student will fre
quently find 3.6035S2 ; a minus sign being written over the characteristic,
to denote that it alone is negative, the mantissa being always positive. The
objection to this notation is the inconvenience of using numbers partly
positive and partly negative.
PROPERTIES OF LOGARITHMS.
452. In any system the logarithm of unity is zero.
For, since a = 1, for any value of a, = log a 1.
453. In any system the logarithm of the base itself is
unity.
For, since a 1 = a, for any value of a, 1 = log„ a.
LOGARITHMS. 343
454. hi any system, whose base is greater than unity, the
logarithm of zero is minus infinity.
For, since or 00 = —  = — = 0, — cc = log a 0.
a go
If the base is less than unity, the logarithm of is + cc .
455. In any system the logarithm of the product of any
number of factors is equal to the sum of the logarithms of
those factors.
Assume the equations,
 "H whence, by Art. 444, [ x ~~ !° g "
Multiplying, a x X cC> — m n, or a x + \ — m n
Whence, x + y = log a m n
Substituting values of x and y,
log a m n — log a m + log a n.
If there are three factors, m, n, andj?,
log a m np = log a (m n Xp) = (Art. 455) log a m n + \og a p
= log a m + log a n + log a p.
An extension of this 'method will prove the theorem for any
number of factors.
By the application of this theorem, we may find the loga
rithm of a number, provided we know the logarithm of each
of its factors. For example, given log 2 = 0.301030, log 3 =
0.477121, required log 72.
log 72 = log (2 x 2 x 2 x 3 X 3)
= log 2 + log 2 + log 2 + log 3 + log 3
= 3 x log 2 + 2 x log 3
= 0.903090 + 0.954242 = 1.857332, Ans.
344 ALGEBRA.
EXAMPLES.
Given log 2 = 0.301030, log 3 = 0.477121, log 7 = 0.845098,
calculate :
•
1. log 48. 4. log 98. 7. log 1G8. 10. log 3087.
2. log 441. 5. log 84. 8. log 7056. 11. log 15552.
3. log 56. 6. log 567. 9. log 504. 12. log 14406.
456. In any system the logarithm of a fraction is equal
to the logarithm of the numerator minus the logarithm of the
denominator.
Assume the equations,
a x = m) i f x = log a m
„ > whence, { , oa
ay — n J (y = \og a n
Dividing,
ar m , m
— = — , or a x ~ v = —
a^ n n
Whence,
xy = \og a 
Substituting
values of x and y,
loga — = log a m — log a n,
IV
By this theorem, a logarithm being given, we may derive
certain others from it. For instance, if we know log 2 =
0.301030, then
log 5 = log ^ = i g io  log 2 = 1.  0.301030 = 0.698970.
41
EXAMPLES.
Given log 2 = 0.301030, log 3 = 0.477121, log 7 = 0.845098,
calculate :
LOGARITHMS. 345
1. log 15. 4. log 175. 7. logTf
2. log 125. 5. log 3i. 8. log—,
10
T
3. log—. 6. loglH. 9. log5£.
457. In any system the logarithm of any power of a
quantity is equal to the logarithm of the quantity, multiplied
by the exponent of the power.
Assume the equation,
a x = in, whence, x = log a m
Eaising both members of the assumed equation to the^th
power,
(a x )P = mP, or aP x = mP
Whence, px = log„ mP
Substituting the value of x,
log a mP=p\og a m.
458. In any system the logarithm of any root of a quan
tity is equal to the logarithm of the quantity, divided by the
index of the root.
For, log a v' m = log a (m7) = (Art. 457)  log a m.
459. In the common system, the mantissa; of the loga
rithms of all numbers having the same sequence of figures
will be the same.
For example, suppose we know that log 3.053 = .484727.
Then,log30.53=log(3.053xl0)=log3.053+logl0=.484727
+ 1 = 1.484727.
Also, log 30530 = log (3.053 x 10000) = log 3.053 + log 10000
= .484727 + 4 = 4.484727.
346 ALGEBRA.
Again, log.03053=log (^~J=log3.053logl00=.484727
2= .484727 + 810 = 8.48472710.
It is clear, then, that if a number he multiplied or divided hy
any integral power of 10, thereby producing another number
having the same sequence of figures, the mantissse of their
logarithms will be the same.
Or, to illustrate, if log 3.053 = .484727,
then, log 30.53 = 1.484727 log .3053 = 9.48472?  10
log 305.3 = 2.484727 log .03053 = 8.484727  10
log 3053. = 3.484727 log .003053 = 7.484727  10
etc. etc.
We may now see the reason why, as stated in Art. 450, only
the mantissa? are given in the table ; for if we wish to find the
logarithm of any number, we have only to find the mantissa
of the sequence of figures composing it from the table, and can
prefix the proper characteristic, depending on the position of
the decimal point, in accordance with the rules stated in Art.
451. This property of logarithms is only enjoyed by the com
mon system, and constitutes its superiority over all others.
460. Given the logarithm, of a quantity to a certain base,
to calculate the logarithm of the same quantity to any other
base.
Assume the equations,
a x = m) i (x = log, m
,,. > whence, < , oa
O'J =m) ' \y = log 6 m
From the assumed equations, a x = by
l I £
Hence, (a*)?/ = (tity, or av = b
Whence,  = lo£„ b
y
X
°h y
l°g a h
LOGARITHMS. 347
Substituting the values of x and y,
, log„ m
log,, m = . .
log„£
That is, if we know the logarithm of m to a certain base, a,
its logarithm to any other base, b, is found by dividing by the
logarithm of b to the base a.
461. To show that log a 6 X log 6 a = l, for any values oj
a and b.
Assume the equation,
a x = b, whence x = log a b
Taking the  power of both members,
00
III
(a x y = b x , or b x = a
Whence,  = log 6 a
7 x
Therefore, log a b X log 6 a = x X  = 1
462. We append a few examples to illustrate the applica
tions of Arts. 455, 456, 457, and 458.
e
L l0g © d= 4 l0g !' (Art. 457)
=  (log a  log b), (Art. 45G) .
Co
2. log % a *V b = log Q a x m \J b)  log % e, (Art. 456)
= log \f a + log y 7 ^ — log y/ c, (Art. 455)
=  log a\ log b log c, (Art. 458).
n m P
The following are proposed as exercises
a b c
~d~e.
3. log 7— = log a + log b + log c — log d — log e.
348 ALGKBBA.
4. log (J/«xJ 3 X c) =  log « + 3 log & + ^ log a.
5. log^ = r, log2log3.
3 g o o
6. log \7 — =  (2 log a — log b — log c).
V c n
„ . V a b 1 „ _ _ . 1
7  log ^y— =  (log « + log &)  — log c.
8  l°g > = t log a — log & — s log c — 2 log tf.
bc$d 2 4 °
( s I a _ ™ \ 1 to
9  lo S VV ^ + ( c d ) "J = 5 ( lo S «  log ^) + — (logc+ logd)
USE OF THE TABLE.
463. The table (Appendix) gives the mantissa? of the
logarithms of all numbers from 1 to 10000, calculated to six
decimal places. On the first page of the table are the loga
rithms of the numbers between 1 and 100. This table is
added simply for convenience, as the same mantissae are to be
found in the rest of the table.
To find the logarithm of any member consisting of four
figures.
Find, in the column headed N, the first three figures of the
given number. Then the mantissa, (if the required logarithm
will be found in the horizontal line corresponding, in the ver
tical column which lias the fourth figure of the given number
at the top. If only the last four figures of the mantissa are
found, the first two figures may be obtained from the nearest
mantissa above, in the same vertical column, which consists of
six figures. Finally, prefix the proper characteristic (Art.
451).
LOGARITHMS. 349
For example, log 140.8 = 2.148603
log .05837 = 8.766190  10
log 8516. = 3.930236
For a number consisting of one or two figures, use the first
page of the table, which needs no explanation ; for a number
of three figures, look in the column headed 1ST, and take the
mantissa corresponding in tbe column headed 0. For exam
ple, log 94.6 = 1.975891.
464. To find the logarithm of a number of more than four
figures.
For example, let it be required to find log 3296.78.
From the table, we find log 3296 = 3.517987
log 3297 = 3.518119
That is, an increase of one unit in the number produces an
increase of .000132 in the logarithm. Then' evidently an in
crease of .78 unit in the number will produce an increase of
.78 X .000132 in the logarithm = .000103 to the nearest sixth
decimal place;
Therefore, log 3296.78 = log 3296 + .000103
= 3.517987 + .000103 = 3.518090, Ans.
Note. The foregoing method is based upon the assumption that the
differences of logarithms are proportional to the differences of their corre
sponding numbers, which is not strictly correct, but is sufficiently exact
for practical purposes.
We derive the following rule from the above operation :
Find in the table the mantissa of the first four figures,
without regard to tin' position of the decimal point.
Find the difference between this and the mantissa of the
next 7iigher number of four figures ; (called the tabular dif
ference, and to be found in the column headed D on each
page : see Note on page 350.)
Multiply the tubular difference by the rest of the figures of
the given number, with a decimal point before them.
Add the result to the mantissa of the first four figures.
Prefix the proper characteristic.
350 ALGEBRA.
1. Find the logarithm of .02243076.
Mantissa of 2243 = 350829
Tabular difference = 194 lj>
.076 350844
1.164
13.58
Correction = 14.744 = 15 nearly.
Am. 8.35084410.
Note. To find the tabular difference mentally, subtract the last figure
of the mantissa from the last figure of the next larger, and take the nean si
whole number ending in that figure to the number in the column headed
D in the same line. For instance, in finding log .02243076, the last figure
of the mantissa of 2243 is 9, and of the next larger mantissa, 3 ; 9 from 13
leaves 4, and the nearest number ending in 4 to 193, the number in the
column headed D, is 194, the proper tabular difference.
EXAMPLES.
Find the logarithms of the following numbers :
10 < "8 oa "" 1
.110 \ji.
"'^ ^""tTlMg XX
LLJ_U.k/^>Xi
z> ■
2.
.053.
6.
33.0908.
10.
912.2:.."..
3.
51.8.
7.
.0002851.
11.
.870092.
4.
.2956.
8.
65000.63.
12.
7303. Oi 8
5.
1.0274.
9.
.001030741.
13.
.0436927
14. Given log 7.83 = .89376, log 7.84 = .89432; find log
78309.
15. Given log .05229 = 8.71841910, log .05230 = 8.718502
10; find log 52.2938.
16 Given log 315.08 = 2.4984208, log 315.09 =2.4984346;
find log .003150823.
17. Given log 18.84 = 1.275081, log 18.87 = 1.275772 ; find
log .188527.
18. Given log 9.5338 = .9792660, log 9.5342 = .9792843 ;
find log 9534071.
LOGARITHMS. 351
465. To find the number corresponding to a logarithm.
For example, let it be required to find the number whose
logarithm is 3.693845.
Since the characteristic depends only on the position of the
decimal point, and in no way affects the sequence of figures
corresponding, we ought to obtain all of the number corre
sponding, except the decimal point, by considering the man
tissa only. We find in the table the mantissa 693815, of which
the corresponding number is 4941, and the mantissa 693903,
of which the corresponding number is 4 ( .)42.
Tbat is, an increase of 88 in the mantissa produces an in
crease of one unit in the number corresponding. Hence, an
increase of 30 in the mantissa will produce an increase of §§ of
a unit in the number, or .34 nearly. Therefore,
Number corresponding = 4941 + 34 = 4941.34, Ans.
We base the following rule on the above operation :
Find in the table the next less mantissa, the four figures
corresponding, and the tabular difference.
Subtract the next less mantissa from the given, mantissa.
Divide the remainder bg the tabular difference / (the quo
tient in general cannot be depended upon to more than two
decimal places.)
Annex all of the quotient except the decimal point to the
first four figures of the number.
Point off.
Note. The rules for pointing off are the reverse of tire rules for charac
teristic given in Art. 451 :
I. If — 10 is not written after the mantissa, add 1 to the
characteristic, giving the number of figures to the left of the
decimal point.
II. If — 10 is written after the mantissa, subtract the
characteynstic from 9 ; giving the number of zeros to be
placed between the decimal point and first figure.
352 ALGEBRA.
1. Find the number whose logarithm is 7.950185 — 10.
950185
Next less mantissa =950170; four figures corresponding =8916.
Tabular difference =49) 15.00 (.31 nearly.
147
~30
Therefore, number corresponding = .00891631, Ans.
EXAMPLES.
Find the numbers corresponding to the following :
2. 1.880814. 6. 8.04489110. 10. 0.990191.
3. 9.47041010. 7. 2.270293. 11. 7.11565810.
4. 0.820204. 8. 935006410. 12. 8.53500310.
5. 4.745126. 9. 3.000027. 13. 1.670180.
14. Given log 113 = 2.05308, log 114 = 2.05690 ; find num
ber corresponding to 1.05411.
15. Given log .08630 = 8.936011  10, log .08631 = 8.936061
— 10 ; find number corresponding to 0.936049.
16. Given log 2.0702 = .3160123, log 2.0703 = .3160333 ;
find number corresponding to 9.3160138 — 10.
17. Given log 548 3 = 2.739018, log 548.9 = 2.739493 ; find
number corresponding to 7.739416 — 10.
18. Given log 7.3488 = .8662164, log 7.3492 = .8662401 ;
find number corresponding to 2.8662350.
466. In the application of Arts. 455, 456, 457, and 458, we
have to perform the operations of Addition, Subtraction, Mul
tiplication, and Division with logarithms. As some of the
problems which may arise arc peculiar, wo give a few hints as
to their solution, which will be found of service.
1. Addition. If, in the sum, — 10, —20, —30, etc., are
written after the mantissa, and the characteristic standing be
LOGARITHMS. 353
fore the mantissa is greater than 9, subtract from both parts
of the logarithm such a multiple of 10 as will make the charac
teristic before the mantissa less than 10.
For example, 13.354802  10 should be changed to 3.354802 ;
28.964316  30 should be changed to 8.964316  10 ; etc.
2. Subtraction. In subtracting a larger logarithm from
a smaller, or in subtracting a negative logarithm from a posi
tive, the characteristic of the minuend should be increased by
10, — 10 being written after the mantissa to compensate.
For example, to subtract 3.121468 from 2.503964, we write
the minuend in the form 12.503964 — 10 ; subtracting from
this 3.121468, we have as a result 9.382496 — 10.
To subtract 9.635321 — 10 from 9.583427  10, we write
the minuend in the form 19.583427 — 20 ; subtracting from
this 9.635321  10, we have as a result 9.948106  10.
3. Multiplication. The hint already given for reducing
the result of Addition, applies with equal force to Multiplication.
To multiply a logarithm by a fraction, multiply first b} r the
numerator, and divide the result by the denominator.
4. Division. In dividing a negative logarithm, add to
both parts of the logarithm such a multiple of 10 as will make
the quantity after the mantissa exactly divisible by the divisor,
with — 10 as the quotient.
For example, to divide 7.402938 — 10 by 6, we add 50 to
both parts of the logarithm, giving 57.402938 — 60. Dividing
this by 6, we have as a result 9.567156 — 10.
EXAMPLES.
1. Add 9.096004  10, 4.581726, and 8.447510  10.
2. Add 7.196070  10, 8.822209  10, and 2.205683.
3. Subtract 0.659321 from 0.511490.
4. Subtract 7.901338  10 from 1.009800;
5. Subtract 9.156243  10 from 8.750404  10.
354 ALGEBRA.
6. Multiply 9.105107  10 by 3.
7. Divide 8.452G33  10 by 4.
8. Divide 9.670392  10 by 11.
9. Multiply 9.6G8311  10 by ?.
SOLUTIONS OF ARITHMETICAL PROBLEMS BY
LOGARITHMS.
467. In finding the value of any arithmetical quantity by
logarithms, we first find the logarithm of the quantity, as in
Art. 462, by the aid of the table, and then find the number
corresponding to the result.
1. Find the value of .0631 X 7.208 X 512.72.
By Art. 455, log (.0631 x 7.208 x 512.72) = log .0631
+ log 7.208 + log 512.72
log .0631= 8.80002910
log 7.208= 0.857815
log 512.72= 2.709880
Adding, .. log of Ans. = 12.367724  10
= 2.367724 (Art. 466, 1)
Number corresponding to 2.367724 = 233.197, Ans.
„. , „ . . 3368.52
2. Find the value of ^^g.
log HJIJH = log 3368.52  log 7980.04
log 3368.52 = 13.527439  10 (Art. 466, 2)
log 7980.04= 3.902005
Subtracting, .. log of Ans. = 9.625434—10
Number corresponding =.422118, Ans.
LOGARITHMS. 355
3. Find the value of (.0980937) 5 .
log (.0980937) 5 = 5 x log .0980937
log .0980937 = 8.991641  10
5
Multiplying, .. log of Ans. = 44.958205  50
= 4.95820510
Number corresponding = .0000090825, Ans.
4. Find the value of ^2.36015.
log ^ 2.3601 5 = * log 2.36015
log 2.36015 = 0.372940
Dividing by 7, .. log of Ans. = 0.053277
Number corresponding = 1.13052, Ans.
2 v^5
5. Find the value of — — .
3*
log $  ] °g 2 + I log 55 log 3
o
log 2 = 0.301030
log 5 = 0.698970 ; divide by 3 = 0.232990
log 3 = 0.477121 0.534020
Multiply by 5, = 2.385605 ; divide by 6 = 0.397601
Subtracting, .. log of Ans. = 0.136419 •
Number corresponding = 1.36905, Ans.
Note. The work of the next two examples will be exhibited in the
customary form, the — 10's being omitted after the mantissse. See Art. 451.
6. Find the value of ^.00003591.
356 ALGEBRA.
log {/ .00003591 = ^ log .00003591
log .00003591 = 5.555215
7)5.555215
log of Ans. Z 9.365031 (Art. 466, 4)
Ans. = .231756.
m n ", i i r // 032956 \
7. Find the value of W ( 
7.96183/'
los \J (^SiH) = \ ^  03295G " log 7,96183)
log .032956 = 8.517934
log 7.96183 = 0.901013
2)7.616921
log of Ans. = 8.808460
Ans. = .0643369.
Note. In computations by logarithms, negative quantities are used as
if they were positive ; the sign of the result being determined irrespective
of the logarithmic work.
EXAMPLES.
468. Calculate; by logarithms, the values of the following :
1. 9.23841 x .00369822. 5. ^3.
*
3.70963 x 286.512 g ,g
1633.72 * ' V
3. (23.846t) 8 . 7. ^5.
4. ( .0009296S7)*. 8. ^.0042937.
LOGARITHMS.
357
18
9. V 6829.586.
112
10. (1.05624)
11. ( .0020001G)i£.
12. 2? x ( 3)*.
13.
14.
3
5 T
(2)*
3^
( 4) §
15. m
ii
16. V 7239.812.
17. V .00230508.
19.
35
113
/ .0872635 U
\ .132088 / "
«. i/\
22. i>
23
21
13*
24. f2x('3xf4.
// 3258.826 \
V V 49309.8 ) '
25
/  31.6259 W
' V 429.0162
27 _ (625.343)
(.732465)
t
28.
29.
30.
V .000128883
y. 000827606*
(_ .746892) ^
 (.234521)^
ty .00730007
" *
(.682913) ^
18. V .000009506694. 31.
y 5.95463 x V 61.1998
V 298.5434
32. (538.217 x .000596899)^.
33. 
304.698 x .9026137
.00776129 X 16923.24
34. (18.9503) 11 x (.213675) 14 .
358 ALGEBRA.
A Orro I Q
35. V 3734.89 x .00001108184.
36. (2.03172)* x (.712719)*
y .00819323 x (.0628513) *
 .9834171 "
6/^T7T7777^T .8/
38. \/.035 x V .626671 X VM721033.
EXPONENTIAL EQUATIONS.
469. An Exponential Equation is one in which the un
known quantity occurs as an exponent.
To solve an equation of this form, take the logarithms of
hoth members according to Art. 457'; the result will be an
equation which can be solved by ordinary algebraic methods.
1. Given 31* = 23 ; find the value of x.
Taking the logarithms of both members,
log (31*) = log 23
or, by Art. 457, x log 31 = log 23
mi lo S 23 1361728 MWVr  ,
Whence, X = ^ = ^—^ = .91307 < , Ans.
The value of the fraction '.^~^ may be obtained bv di
1.491362 J J
vision, or better by logarithms, as in Art. 468.
2. Given .2* = 3 ; find the value of x.
Taking the logarithms of both members,
x log .2 = log 3
log 3 .477121 .477121
\Y hence, x =
' log .2 9.301030 — 10 .698970
We may find the value of the fraction by logarithms exactly
as if it were positive, and prefix a — sign to the result. Thus.
LOGARITHMS. 359
log .477121 = 9.678628  10.
log .698970 = 9.814458  10
Subtracting, = 9.834170  10
Number corresponding = .682606
Therefore, x = — .682606, Ans.
EXAMPLES.
Solve the following equations :
3. II 1 = 3. 5. 13* = .281. 7. 5*~ 8 = 8 2 * +1 .
4. .3 r = .S. 6. .703* = 1.09604. 8. 23 3 * + 5 = 31 2 * 3 .
APPLICATION OF LOGARITHMS TO PROBLEMS IN
COMPOUND INTEREST.
470. Let P = the principal, expressed in dollars.
Let t = the interval of time during which simple interest
is calculated, expressed in years and fractions of a }*ear. For
instance, if the interest is compounded annually, t = 1 ; if
semiannually, t =  ; etc.
Let P = the interest of one dollar for the time t.
Let n = the number of years.
Let Ai, A 2 , A 3 , be the amounts at the ends of the 1st,
2d, 3d, intervals.
Let A be the amount at the end of n years.
Then A l = P + PP = P(l + P)
A 2 = A 1 + A 1 B = A 1 (1 + R)
=p (i + p) (i + p) = p (i + py
A 3 = A, + A 2 P = A 2 (1 + P)
= p(i + py 2 (i + p) = p(i + py
300 ALGEBRA.
71
As there are  intervals, the amount at the end of the last,
v
according to the law observed above,
A =* P (1 + Sp.
1. Given P, t, R, and n, to find A.
n
As A = P (1 + R) *., we have by logarithms,
log A = log P (1 + Rp = log P + log (1 + Rp
71
= log P +  log (1 + R).
h
Example. What will be the amount of $7,325.67 for 3
years 9 months at 7 per cent compound interest, the interest
being compounded quarterly ?
Here P = 7325.67, t = j , R — .0175, n = 3f,  == 15.
log P = 3.864848
i
log (1 + R) = 0.007534 ; multiply by 15 = 0.113010
Adding, . •. log of A = 3.977858
Number corresponding, A = $ 9502.93, Ans.
2. Given t, R, n, and A, to find P.
n A
As A = P (1 + R)t , .. P = ; or, by logarithms,
(1 + R)~
/log P = log A — log (1 + Rp = log A — " log (1 + R).
Example. What sum of money will amount to $ 1 ,76« \.5B at
5 per cent compound interest in 3 years, the interest being
compounded semiannually ?
LOGARITHMS. 361
1 n
Here t = % , B = .025, » = 3, A = 1763.55,  = 6.
log ^ = 3.246388
log (1 + P) = 0.010724; multiply by 6 = 0.064344
Subtracting, .. log P = 3.182044
Number corresponding = f> 1520.70, Ans.
3. Given P, t, P, and A, to find n.
In Art. 470, 1, we sbowed that
log^ = logP + log(l + P)
1/
.•.^log(l + P)=log^logP
£ (log A — log P)
* • l ~ log (1 + 5) •
Example. In bow many years will $300.00 amount to
8 400.00 at 6 per cent compound interest, the interest being
compounded quarterly ?
Here P = 300, t =  , P = .015, A = 400.
log 400  log 300 2.602060  2.477121 .124039
.'.71 =
4 log 1.015 ~ 4 x .006466 .025864
= 4.83 years, Ans.
4. Given P, t, n, and A, to find P.
n
We sbowed, in Art. 470, 3, that  log(l + P) = log^logP
V
1 /1 , T>\ log^ — logP
••• log (1 + P) ■ •
Example. If $ 500.00 at compound interest amounts to
$689.26 in 6 years and 6 months, the interest being com
pounded semiannually, what is tbe rate per cent per annum '?
G2
ALGEBRA.
1 n
Here P = 500, t = 7> ,n = 6h,A = 689.26,  = 13.
z t
, , ™ log 689.26 log 500
13
log 689.26 = 2.838383
log 500 = 2.698970
Subtracting,
= 0.139413
Dividing by 13, .. log (1 + B) = 0.010724
Number corresponding = 1.025 = 1 + II, or R = .025.
That is, one dollar gains $ .025 semiannually ; or the rate is
5 per cent per annum.
EXPONENTIAL AND LOGARITHMIC SERIES.
471. We know that for any values of n and x,
1 +
x / 1 \ n
= ( 1 + n,
Expanding by the Binomial Theorem, we obtain
1 n (n  1) 1 n (n  1) (n  2) 1
1 + n  H r?i 5 H rr; ? +
w
w."
[3
?r
. 1 v x (nr—1) 1 ?? x(nx— V)(nx— 2) 1
w 2 ?r 3 ?r
+
or,
!_! (ll)(! 2 )
T77 T^ +
L3
x \x x \x
V n I n
= 1 + x H r^ +
11
\1
; +
LOGARITHMS. 363
This is true for all values of n ; hence, it is true however
large n may he. Suppose n to he indefinitely increased. Then
1 2
the limiting values of the fractions  , — , etc., are (Art. 210).
n n
Hence, at the limit, we have,
1 1
1 + 1 + 777 + T7V +
f^ + Sr +
[3_ J ' ' [2 ' [3
The series in the bracket we denote by e ; hence,
2 3
x x
<? = ! + * +  +  +
472. To expand a x in poivers of x.
Let a = e m ; whence (Art. 444), 7ft = log c ft.
m 2 a? 2 ra 3 x 3
Then a* = e ra x = (Art. 471) 1 + m a; + y^ + r^ +
Substituting the value of m,
V 2 /I \ *\
a* = 1 + (l0g e ft) X + (log, ft) 2 — + (log e ft) 3 j^ +
This result is called the Exponential Theorem.
473. The system of logarithms which has e for its base,
is called the Napierian System, from Napier, the inventor of
logarithms. The value of e may be easily calculated from the
series of Art. 471, and will be found to be 2.7182818
474. To expand log,, (1 + x) in poivers ofx.
a* = {l+(al)}* = l + x(a l l).+ X(x ~ 1) (aiy
+ g (sl)(s2) +
= l + x {(ft 1)   ( ^ + ^^ } + terms con
taining x 2 , x s } etc.
364 ALGEBRA.
But (Art. 472), a x = 1 + x (log e a) + terms containing x%
As the two values of a x are equal for all values of x, by the
Theorem of Undetermined Coefficients the coefficients of x in
the two expressions are equal; hence,
log e a=(a — l) ^ 1 3
Putting a = 1 + x, and therefore a — 1 = x, we obtain
Syt" /ytO
log e (l + «)=» — — + — —
Note. This formula might be used to calculate Napierian logarithms ;
but unless x is a very small fraction, the series in the second number is
either divergent or converges very slowly, and hence is useless in most
cases.
475. To obtain a more convenient formula for calculating
the Napierian logarithm of a number.
X 2 X s X* X 5
By Art. 4,4, log e (1 + x) = x — y + y — y + y —
put X = — X,
X 2 X 3 X* X s
..log e (l — x)=—x
Subtracting,
2 x 3 2x h
• . log e (1 + x)  log, (lx) = 2x+ — + ir +
or, by Art. 456, log,, f j
1 + x\ c ( x 3 x 5
> = 2 [x+ 7r + T r +
Let x =
x) 3 5
1
1 +
2n + l
1
l+x_ 2n + l _ 2n + 1 + 1 _ 2n + 2 __ n + l
l—x~~7 ~T~ ~2n + l — l~ 2 u n
2n + l
LOGARITHMS. 365
Substituting, .\ log, p— J = lo ge (n + 1) — log e »
~~ \2n + 1^3(2n + iy ^ 5 (2 n + l) 3 T " /
■■■^g e (n+l)=\o ge n+2(^ + 3 ^ +1) ,+ 5{:Jn+1)b + )
476. To calculate log e 2, put n = 1 in the formula of Art.
475.
/ 1 1 1 \
...log e 2 = log e H2^2 TI +3 (2 + 1) s+5^2q:i)5+ j
or, since log e 1 = 0,
/l 1 1 1 1 1
log. 2 = 2 ^3 + 81 + 12i5 + 15309 + 177147 + 1948617" 1 """
= 2 (.3333333 + .0123457 + .0008230 + .0000653
+ .0000056 + .0000005 + )
= 2 x .3405734 = .6931468 = .693147, correct to the
sixth decimal place.
From log,, 2, we may calculate log e 3 ; and so on. We shall
find log e 10 = 2.302585.
477. Tn calculate the common logarithm of a number from
its Napierian logarithm.
By Art. 460, changing b to 10, and a to e, we obtain
"»■ " = feS = &3S3«i b ~ » =  434  0945 X ,0g ' *
For instance, log I0 2 = .4342945 x .693147 = .301030.
The multiplier by which logarithms of any system are de
rived from the Napierian system, is called the modulus of that
system. Hence, .4342945 is the modulus of the common sys
tem.
366 ALGEBRA.
As tables of common logarithms are met with more fre
quently than tables of Napierian, a rule for changing common
logarithms into Napierian may be found convenient.
RULE.
Divide the common logarithm by .4342945.
For example, to find the Napierian logarithm of 586.324,
common log 586.324 = 2.768138
Divide by .4342945, .. Napierian log 586.324 = 6.373873, Ans.
Another method would be to multiply the common logarithm
by 2.302585, the reciprocal of .4342945.
Napierian logarithms are sometimes called hyperbolic loga
rithms, from having been originally derived from the hyper
bola. They are also sometimes called natural logarithms,
from being those which occur first in the investigation of a
method of calculating logarithms. Napierian logarithms are
seldom used in computation, but occur frequently in theoretical
investigations.
ARITHMETICAL COMPLEMENT.
478. The Arithmetical Complement of the logarithm of
any quantity is the logarithm of the reciprocal of that quantity.
For example, if log 4098 = 3.612572, then
ar. co. l#g 4098 = log — — = log 1  log 4098
° 4098 ° °
=  3.612572 = 6.387428  10.
Again, if log .06689 = 8.825361  10, then
ar. co. log .06689 = log ?—  =  (8.825361  10)
= 10  8.825361 = 1.174639.
LOGARITHMS. 367
The following rules will be evident from the preceding
illustrations :
To find the arithmetical complement of a positive loga
rithm, suit nut it from 10, writing — 10 after the mantissa.
To find the arithmetical complement of a negative loga
rithm, subtract that portion of it besides the — 10 from 10.
The only application of this is to exhibit the work of calcu
lation by logarithms in a more compact form in certain cases.
It depends on the' principle that subtracting a logarithm or
adding its arithmetical complement gives the same result.
For, suppose we are to calculate by logarithms.
. a X b . ( . 1 1
l0 «7^ = l0g l ax( ' x c x ( z
= log a + log b + log  + log
1 i 1
c + ] ° S d
= log a f log b + ar. co. log c + ar. co. log d.
That is, the work can be exhibited in the form of the addi
tion of four logarithms, instead of the subtraction of the sum
of two logarithms from the sum of two others. The principle
is only applicable to the case of fractions ; and the rule to be
used is,
Add together the logarithms of the quantities in the numer
ator, and the arithmetical complements of the logarithms of
the quantities in the denominator.
Example. Calculate the value of „ 't, — _1 00 .
1 613.8 x .0* .23
log Qi^xljfn = l0 S 79 ' 23 + lQ g 10  39 + ar " co  Io S 613  8
+ ar. co. log .07723
368 ALGEBRA.
log 79.23= 1.898890
log 10.39 = 1.016616
ar. co. log G13.8 = 7.211973  10
ar. co. log .07723= 1.112211
Adding, ..log of Ans. = 11.239693  10 = 1.239693
Number corresponding = 17.3657, Ans.
Note. The arithmetical complement may be calculated mentally from
the logarithm, by subtracting the last significant figure from 10, and all the
others from 9.
MISCELLANEOUS EXAMPLES.
479. 1. Find log 3 2187. (See Art. 111.)
2. Find log 5 15625.
3. Find the logarithm of rr to the base —2.
* 61
4. Find the logarithm of — to the base 8.
5. Find the characteristic of log 2 183.
6. Find the characteristic of log 5 4203.
7. Given log 2 = .301030, how many digits are there in
2 17 ?
8. Given log 3 = .477121, how many digits are there in
3^?
9. Findlog 13 56. (See Art. 460.)
10. Find log 8 163.
11. Find loggo 411.
12. What sum of money will amount to § 8705.50, in 7
years, at 7 per cent compound interest, the interest being com
pounded annually ?
GENERAL THEORY OF EQUATIONS. 360
13. In how many yours will a sum of money double itself at
6 per cent compound interest, the interest being compounded
semiannually ?
14. What will be the amount of $1000.00 for 38 years
3 months, at 6 per cent compound interest, the interest being
compounded quarterly ?
15. At what rate per cent per annum will $2500.00 amount
to $ 3187.29 in 3 years and 6 months, the interest being com
pounded quarterly ?
16. In bow many years will 8 9681.32 amount to $ 15308.70
at 5 per cent compound interest, the interest being compounded
annually ?
17. Using the table of common logarithms, find the Na
pierian logarithm of 52.9381 (Art. 477).
18. Find the Napierian logarithm of 1325.07.
19 Find the Napierian logarithm of .085623.
20. Find the Napierian logarithm of .342977.
XLII. — GENERAL THEORY OF EQUATIONS.
480. The general form of a complete equation of the nth
degree is
x n \ 1~> x n ~ l + q x n ~' 2 + + t x 2 + u x + v =
"Where n is a positive integer, and the number of terms is n + 1.
The quantities j), q, t, u, v are either positive or nega
tive, integral or fractional ; and the coefficient of x 11 is unity.
481. In reducing an equation to the general form, all the
terms must be transposed to the first member, and arranged
according to the powers of x. If as" has a coefficient, it may
be removed by dividing the equation by that coefficient.
370 ALGEBRA.
482. A Root of an equation is any real or imaginary ex
pression, which, being substituted for its unknown quantity,
satisfies the equation, or makes the first member equal to
(Art. 166).
We assume that every equation has at least one root.
483. An equation of the third degree containing only one
unknown quantity, or one in which the cube is the highest
power of the unknown quantity, is usually called a cubic equa
tion.
484. An equation of the fourth degree containing only one
unknown quantity is usually called a biquadratic equation.
DIVISIBILITY OF EQUATIONS.
485. If a is a root of an equation in the form
x n +qi x n ~ l + q x n ~ 2 + + tx 2 + ux + V — 0,
then the first member is divisible by x — a.
It is evident that the division of the first member by x — a
maybe carried on until x disappears from the remainder. Let
Q represent the quotient, and R the remainder, which is inde
pendent of x ; then the given equation may be made to take
the form
. (x  a) Q + B = 0.
But if x = a, then (x — a) Q = 0, and, consequently,
R = 0;
that is, x — a is a factor of the first member of the given equa
tion, as it is contained in it without a remainder.
486. Conversely, if the first member of am. equation in the
form
*
x n \px n ~ 1 \ qx n ~ 2 + + t x 2 + ux + v =
is divisible by x — a, then a is a root of the equation.
GENERAL THEORY OF EQUATIONS. 371
For, if the first member of the given equation is divisible by
x — a, then tbe equation may be made to take the form
(x — a) Q = ;
and it follows from Art. 330 that a is a root of this equation.
EXAMPLES.
By the method of Art. 486,
1. Prove that 3 is a root of the equation
x 3 — 6 x 2 + 11 x — 6 = 0.
2. Prove that — 1 is a root of the equation x 3 + 1 = 0.
3. Prove that 1 is a root of the equation
x 3 + x 2 — 17 a; + 15 = 0.
4. Prove that — 2 is a root of the equation
a; 4 — 3 x 2 + 4 x + 4 = 0.
5. Prove that 4 is not a root of the equation
x* — 5 x 3 + 5 x 2 + 5 x  6 = 0.
NUMBER OF ROOTS.
487. Every equation of the nth degree, containing but one
unknown quantity, has n roots, and no more.
Let a be a root of the equation
x n +2) x n ~ x + qx n ~ 2 + + tx 2 + ux + v — 0;
then, by Art. 485, the first member is divisible by x — a, and
the equation may be made to take the form
(x — a) (V 1  1 + 2h x n ~ 2 + + «j x + i\) = 0.
The equation may be satisfied by making either factor of
the first member equal to (Art. 330) ; hence,
x • — a =
and x»~ 1 +p 1 x n  i + +u l x + v 1 = 0. (1)
372 ALGEBRA.
But equation (1) must have some root, as b, and may be
placed under the form
(x — b) (x 11  2 +jh x n ~ 3 + + v,x + v,)=0',
which is satisfied by placing either factor of the first member
equal to ; and so on.
Since each of the factors x — a, x — b, etc., contains only
the first power of x, it is evident that the original equation can
be separated into as many such binomial factors as there are
units in the exponent of the highest power of the unknown
quantity, and no more ; that is, into n factors, or
(x — a) (x — b) (x — c) (x — T) = 0.
Hence, by Art. 330, the equation has the n roots a, b, c, /.
Moreover, if the equation had another root, as r, then it
must contain another factor x — r, which is impossible.
488. It should be observed that the n binomial factors of
which the general equation of the nth degree is composed, are
not necessarily unequal j hence, two or more of the roots of an
equation may be equal. Thus, the equation
x * _ g x 2 + 12 x _ 8 =
may be factored so as to take the form
(x2)(x2) (.r2)=0, or (*2) 3 = 0;
and hence the three roots are 2, 2, and 2.
489. It will be readily seen that any equation, one of
whose mots is known, may be depressed to another of the next
lower degree, which shall contain the remaining roots. Hence,
if all the roots of an equation are known excepl two, those
may be obtained from the depressed equation, hy the rules for
quadratics.
1. One root of the equation x s + 2 x 1 — 23 x — 60 = is — 3 ;
what are the others ?
GENERAL THEORY OF EQUATIONS. 373
Dividing x 3 + 2 x~ — 23 x — 60 by x + 3, the given equation
may be put in the form
O + 3)O 2 x20)=0.
Thus, the depressed equation is x 2 — x — 20 = 0.
Solving this by the rules for quadratics, we obtain x = 5 or
— 4 ; which are the remaining roots.
EXAMPLES.
2. One root of the equation x z — 19 x + 30 = is 2 ; what
are the others ?
3. Eequired the three roots of the equation X s = a 3 , or
x 3 — a 3 — 0.
4. One root of the equation x 3 + x 2 — 16 x + 20 = is — 5 ;
reqxiired the remaining roots.
5. Two roots of the equation x A — 3 x 3 — 14 x" + 48 x — 32 =
are 1 and 2 ; required the remaining roots.
6. One root of the equation x 4 — 7 x 3 + 3 x + 3 = is 1 ;
what equation contains the remaining roots ?
7. One root of the equation 6 x 3 — x~ — 32 x + 20 = is 2 ;
what are the others ?
8. Two roots of the equation 20 x*  169 x 3 + 192 x~ + 97 x
— 140 = are 1 and 7 : what are the others ?
5
FORMATION OF EQUATIONS.
490. An equation having any given roots may be formed
by subtracting each root from the unknown quantity, and pla
cing the product of these binomial factors equal to 0.
For it is evident, from principles already established, that
an equation having the n roots a, b, c, I may be written
in the form
(x — a) (x — b) (x — c) (x — I) = 0.
374 ALGEBRA.
After performing the multiplication indicated, the equation
will assume the form .
x n +2?x n ~ 1 + qx n ~ 2 + + tx 2 + ux + v = 0.
(Compare Art. 329.)
1. Form the equation whose roots are 1, 2, and — 4.
Result, (x — 1) (cc — 2) (cc + 4) =
or, x 3 + x 2 — 10 x + 8 = 0.
EXAMPLES.
Form the equations whose roots are :
2.  1,  3, and  5. 6. 1, 2, 3, and 4.
3. 5,  2, and  3. 7. 4, 4, and 5.
4. 1,  , and  . 8. 0,  1, 3, and 4.
5. ± 1 and ±2. 9.  5 ? ,  2, and ?
4 3
COMPOSITION OF COEFFICIENTS.
491. The coefficient of the second term of an equation of the
nth degree in its general form is the sum of all the roots with
their signs changed; that of the third term is the sum of their
products, taken two and two ; that of the fourth term is the
sum of their 'products, taken three and three with their .signs
changed, etc. ; and the last term is the product of all the roots
with their signs changed.
For, resuming the equation
(x '— a) (x — b)(x — c) (x — k) (x — l)=0,
if we perform the multiplication indicated, we obtain
(x — a) (x — b) = x~ — (a + b) x + a b,
(x—a)(x — b) (x—c) = x 3 —(a + b+c)x+(ab + ac+bc)x—abc,
GENERAL THEORY OF EQUATIONS. 375
and so on. When n factors have heen multiplied, the coeffi
cients of the general equation become
jp = — a — b —c— — k — I
q = ab + ac\bc\ + kl
r = — a b c — ab d — a c d — — ikl
v = ±a b c kl
which corresponds with the enunciation of the proposition ;
the upper sign of the value of v being taken when n is even,
and the lower sign when n is odd.
492. If ^=0, that is, if the second term of an equation
be wanting, the sum of the roots will be 0.
If v = 0, that is, if the absolute term of an equation be want
ing, at least one root must be 0.
493. Every rational root of an equation is a divisor of the
last term.
494. When all the roots of an equation but two are known,
the coefficient of the second term of the depressed equation
(Art. 489) can be found by subtracting the sum of the known
roots, with their signs changed, from the coefficient of the
second term of the original equation. The absolute term of
the depressed equation can be found by dividing the absolute
term of the original equation by the product of the known
roots with their signs changed.
EXAMPLES.
Find the sum and product of the roots in the following :
1. a; 3 7a;+6 = 0. 2. 2 x* 5 x s  17 x 2 + 14 x + 24 = 0.
In the following example obtain the depressed equation by
the method of Art. 494 :
3. Two roots of the equation x 4 — 5 x s — 2 x 2 + 12 x + 8 =
are 2 and — 1 ; what are the others ?
376 ALGEBRA.
FRACTIONAL ROOTS.
495. An equation whose coefficients are all integral, the
coefficient of the first term being unity, cannot have a
rational fraction as a root.
If possible, let , a rational fraction in its lowest terms, be
a root of the equation
x n + px n ~ l + qx n ~ 2 + + tx 2 + UX + v = 0,
where p, q, , t, u, v are integral. Then
a\ n faV' 1 fa\ n  2 ( a\ 2 fa
Multiplying through by lf~ l , and transposing,
 n = — ( 2 ,a n  1 +qa n  2 b + + ta 2 b n ~ s + ua u n ~ 2 + vb n ~ l ).
Now, as  is in its lowest terms, a and b can have no com
mon divisor ; therefore a n and b can have no common divisor ;
a"
hence — is in its lowest terms. Thus, we have a fraction in
o
its lowest terms equal to an entire quantity, which is impossi
ble. Therefore no root of the equation can be a rational
fraction.
Note. The equation may have an irrational fraction as a root, such as
' for example. Such a root, whose value can only be expressed
4
approximately by a decimal fraction, is called incommensurable.
IMAGINARY ROOTS.
496. If the coefficients of an equation be real quantities,
imaginary roots enter it by pairs, if at all.
Suppose a + b ^ — 1 to be a root of the equation
X n + p X" ~ l + q X n ~ 2 + + t X 2 + U X + V = 0.
GENERAL THEORY OF EQUATIONS. 377
Substituting a + b y/— 1 for x, and developing each expres
sion by the Binomial Theorem, all the odd terms of each series
will contain either powers of a, or even powers of b y/ — 1, and
are therefore real ; while all the even terms contain the odd
powers of b y — 1, and are therefore imaginary. Representing
the sum of all the real quantities by P, and the sum of all
the imaginary quantities by Q y — 1, we have
P+ £y/l=0.
This equation can be true only when both P and Q equal 0.
If we now substitute a — b y/ — 1 for x, we find that the
series differ from the former only in having their even or
imaginary terms negative. Hence, we obtain as the first
member
PQ^l,
which must be equal to 0, for we have already shown that both
P and Q equal 0. Thus, a — b y/— 1 satisfies the equation.
Similarly, we may show that, if b y/ — 1 is a root of the equa
tion, then will — b y/— 1 also be a root of the equation.
497. The product of a pair of imaginary quantities is
always positive. Thus,
(a + 6y/l) (ab^l) = a 2 + b
and y/^l) ( b y/^I) = b\
TRANSFORMATION OF EQUATIONS.
498. To transform an equation into another which shall
have the same roots with contrary signs.
Let the given equation be
x n + p x n ~ 1 + q x n ~ 2 + + t x 2 + u x + v = 0.
378 ALGEBRA.
Put x = — y; then whatever value x may have, y will have
the same value with its sign changed. The equation now
becomes
{y) n +p(y) n  l + q(y) n  2 + + t(y) 2 + u{ y )
+ v = o.
If 11 is even, the first term is positive, second term nega
tive, and so on ; and the equation may be written
yn_ p yU\ + (1 yn~1_ + f y'2 _ ^ y _J_ y _ Q Qj
If n is odd, the first term is negative, second term positive,
and so on ; hence, changing all signs, we write the equation
y" —v y n ~ 1 + 1 y n ~ 2 — — ^ y 1 + u y~ v — o. (2)
From (1) and (2) it is evident that to effect the desired
transformation we have simply to change the signs of the
alternate terms, beginning with the second.
Note. The preceding rule assumes that the given equation is complete
(Art. 300) ; if it be incomplete, any missing term must be put in with
zero as a coefficient.
1. Transform the equation x 3 — 7x + 6 = into another
which shall have the same roots with contrary signs.
We may write the equation x 3 + . ar — 7 x + 6 = 0.
Applying the rule,
x s _ o . x 2 — 7 x — 6 = 0, or x z — 7 x — 6 = 0, Ans.
EXAMPLES.
Transform the following equations into others which shall
have the same roots with contrary signs:
2. x *  2 x s + x  1.32 = 0. 3. x 5 3x 2 + 8 = 0.
499. To transform an equation into another whose roots
shall be some multiple of those of the first.
Let the given equation he
x n +px n  1 + qx n ~ 2 + + tx' 2 + ux + v = 0.
GENERAL THEORY OF EQUATIONS. 379
v
Put x = — ; then whatever value x may have, y will have
m
a value m times as great. The equation now becomes
(i)r + jjLY\Ji\"\ +,(£)*+.(»)+.=*
Multiplying through by ra n , we have
y n +pmy n ~ 1 +qm 2 y n ~ 2 + + tm n ~ 2 y 2 + um n ~ 1 y + vm" = 0.
Hence, to effect the desired transformation, multiply the
second term by the given factor, the third term by its square,
and so on.
Similarl}', we may transform an equation into one whose
roots shall be those of the first divided by some quantity.
1. Transform the equation x 3 — 7 x — 6 = into another
whose roots shall be 4 times as great.
The equation may be written, x s f . x 2 — 7 x — 6 = 0.
Then, by the rule,
x 3  4 2 . 7 x  4 3 . 6 = 0, or x 3  112 x  384 = 0, Ans.
EXAMPLES.
2. Transform the equation x z — 2 x 2 + 5 = into another
whose roots shall be 5 times as great.
3 x s
3. Transform the equation x i  — 27 = into another
whose roots shall be one third as great.
500. Ta transform an equation containing fractional
coefficients into another whose coefficients are integral, that of
the first term being unity.
If in Art. 499 we assume m equal to the least common
multiple of the denominators, it will always remove them ;
but often a smaller number can be found which will produce
the same result.
380 ALGEBRA.
1. Transform the equation x s — — — — + — — = into
o So lUo
another whose coefficients shall he integral.
The least common multiple of the denominators is 108 ; so
that one solution would he, by Art. 499,
rf_M8. .M8: + 108^ = 0.
An easier way, however, is as follows ; the denominators
may be written 3, 3 2 X 2' 2 , and 3 3 X 2 2 , so that the multiplier
3 x 2 or 6 will remove them. Hence, by Art. 499, we have
x a_ 6 * a __ 6 3 * +6 s * =0 or x *2x*x + 2 = 0,
ob 106
whose roots are 6 times as great as those of the given
equation.
EXAMPLES.
Transform the following equations into others whose coef
ficients shall be integral :
**'+T~ 7 4=  *. , + reg»= a
av^ + fwb. 5.^5^^ + 5 = 0.
no 4 I
501. To transform an equation into another ivhose roots
shall be the reciprocals of those of the first.
Let the given equation be
x n + p x n ~ 1 + q x n ~  + + t x 2 + u x + v = 0.
Put x =  ; then whatever value x may have, y will be its
v
reciprocal. The equation now becomes
1 p q t u n
y y y y y
GENERAL THEORY OF EQUATIONS. 381
Multiplying through by y", and reversing the order,
vy n + uy n ~ 1 + ty" 2 + + qy 2 +py + 1 = 0.
Dividing through by v,
u , t a „ 7} 1
n + _ y «i +  gf» + + Lf + L l/+ ± =0 .
v « w «; «;
Hence, to effect the transformation, write the coefficients in
reverse order, and then divide by the coefficient of the first
term.
EXAMPLES.
Transform the following equations into others whose roots
shall he the reciprocals of those of the first :
1. a; 3_6x 2 +ll iC 6 = 0. 3. x 3 9x 2 + — i = 0.
7 49
2. x i ~x 3 3x 2 + x + 2 = 0. 4. x 3 4x 2 + 9 = 0.
502. To transform an equation into another whose roots
shall differ from those of the first by a given quantity.
Let the given equation be
s"+ju""' + qx n ~ 2 + + tx* + ux + v = 0. (1)
Put x = y + r, and we have
(ff + r)*+p(t,. + r)— 1 + + u (y + r ) + v=0. (2)
Developing (y + r) n , (y + r)"\ , by the Binomial The
orem, and collecting terms containing like powers of y, we
have an equation of the form
yn +pi ynl + ^ y «1 + + ^ y 2 + ^ y + ^ = Q ( o)
As y=x — r, the roots of (3) are evidently less by r than
those of (1). By putting x = y — r, we shall obtain in the
same way an equation whose roots are greater by r than
those of (1).
503. If n is small, the operation indicated in Art. 502
may be effected with little trouble; but for equations of a
higher degree a less tedious method is better.
382 ALGEBRA.
If in (3) we put y = x — r, we shall have
(x—rf +2h(x — r) n  1 + +u 1 (x — r) + v 1 = 0, (4)
which is, of course, identical with (1), and must reduce to (1)
when developed. If we divide (4) by x — r, we obtain
(x  r)" ~ J +jh  r) n  2 + q i {xr)""+ +u, (5)
as a quotient, with a remainder of v v Dividing (5) by x — r,
we obtain a remainder of u x ; and so on, until we obtain all
the coefficients of (3) as remainders.
Hence, to effect the desired transformation,
Divide the given equation by x — r or x + r> according as
the roots of the transformed equation are to be less or greater
than, those of the first by r, and the remainder will be the
absolute term of the transformed equation. Divide the quo
tient just found by the same divisor, and the remainder will
be the coefficient of the last term but one of the transformed
equation ; and so on.
504. 1. Transform the equation x 3 + 3 x 2 — 4^ + 1 =
into one whose roots shall be greater by 1.
Using the method of Art. 502, put x = y — 1.
Then, (yiy + 3(y~iy4(yl) + l = 0,
or, t/ 3  3 y 2 + 3 7/  1 + 3 /  6 y + 3  4 y + 4 + 1 = 0,
or, if — 7 y + 7 = 0, Ans.
EXAMPLES.
2. Transform the equation x 3 — x — 6 = into one whose
roots shall be less by 8.
3. Transform the equation x* f 6 x 3 — x 1 — 5 x — 1 =
into one whose roots shall be greater by 3.
505. To transform a complete equation into one tchos*
second term shall be wanting.
GENERAL THEORY OF EQUATIONS. 383
The coefficient of y" 1 in (2), Art. 502, is n r + p. Hence,
in (3), p y = n r + p. To make ^ = 0, it is only necessary to
make nr + p =0, or r = — — ; hence, to effect the desired
n
V
transformation, put x = y —  ; that is, put x equal to y }
minus the coefficient of the second term of the given equation
divided by the degree of the equation.
1. Transform the equation x 3 — G x~ j 9 x — G = into
another whose second term shall be wanting.
Here p = — 6, n = 3 ; then, put x = y — — — = y + 2.
Result, (y + 2) 3  6 (y + 2) 2 + 9 (y + 2)  6 = 0,
or, tf + 6y* + 12y + 86y*24:y24:+9y +186 = 0,
or, y 3 — 3 y — 4 = 0, Ans.
EXAMPLES.
Transform the following equations into others whose second
terms shall be wanting :
2. x 2 px + q = 0. 4. x 3 + 6x 2 3x + 4: = 0.
3. x s + x 2 + 4 = 0. 5. x 4  4 x 3  5 x — 1 = 0.
DESCARTES' RULE OF SIGNS.
506. A Permanence of sign occurs when two successive
terms of a series have the same sign.
A Variation of sign occurs when two successive terms of a
series have contrary signs.
DESCARTES' RULE.
507. A complete equation cannot have a greater number
of positive roots than it has variations of sign, nor a greater
number of negative roots than it has permanences of sign.
384 ALGEBRA.
Let any complete equation have the following signs :
+ + +  +  +
in which there are three permanences and five variations.
If we introduce a new positive root a, we multiply this by
x — a (Art. 490). Writing only the signs which occur in the
operation, we have
+ + +  +
— 1
+ —
+ + +  +
 + —
+
 + — + +
+ ± ± — + — + — ± +
123456789 10
a double sign being placed wherever the sign of a term is
ambiguous.
However the double signs are taken, there must be at least
one variation between 1 and 4, and one between 8 and 10,
and there are evidently four between 4 and 8 ; or in all there
are at least six variations in the result. As in the original
equation there were five variations, the introduction of a
positive root has caused at least one additional variation ; and
as this is true of any positive root, there must be at least as
many variations of sign as there are positive roots.
Similarly, hy introducing the factor x + a, we may show
that there are at least as many permanences of sign as there
are negative roots. ,
If the equation is incomplete, any missing term must be
supplied with ± as its coefficient before applying Descartes'
Rule.
508. In any complete equation, the sum of the number
of permanences and variations is equal to the number of terms
less one, or is equal to the degree of the equation (Art. 480).
Hence, when the roots are all real, the number of positive
roots is equal to the number of variations, and the number of
negative roots is equal to the number of permanences (Art. 4S7).
GENERAL THEORY OF EQUATIONS. 385
A complete equation whose terms are all positive can have
no positive root ; and one whose terms are alternately positive
and negative can have no negative root.
509. In an incomplete equation, imaginary roots may
sometimes he discovered hy means of the douhle sign of in
the missing terms. Thus, in the equation
x 3 + x 2 ± x + 4 =
if we take the upper sign, there is no variation, and conse
quently no positive root; if we take the lower sign, there is
hut one permanence, and hence but one negative root. There
fore, as the equation has three roots (Art. 487), two of them
must he imaginary.
In general, whenever the term which precedes a missing
term has the same sign as that which follows, the equation
must have imaginary roots ; where it has the opposite sign,
the equation may or may not have imaginary roots, hut
Descartes' Hule does not detect them. If two or more suc
cessive terms of an equation he wanting, there must be imagi
nary roots.
Note. In all applications of Descartes' Rule, the equation must con
tain a term independent of x, that is, no root must be equal to zero (Art.
330) ; for a zero root cannot be considered as either positive or negative.
EXAMPLES.
510. The roots of the following equations being all real,
determine their si cms :
*&'
1. x 8 3x2 = 0. 3. a; 8 7 a; 2 + 36 = 0.
2. .x 3 10a; + 3 = 0. 4. x i 2x 3 13x 2 + 38*24 = 0.
5. What are the signs of the roots of the equation x 3 + x 1
4 = 0?
DERIVED POLYNOMIALS.
511. If we take the polynomial
a x n + b x n ~ 1 + c x n ~ 2 +
386 ALGEBRA.
and multiply each term by the exponent of a; in that term,
and then diminish the exponent by 1, the result
n a x" ~ l + (n — 1) b x H ~ 2 + Qi — 2) c x n ~ 3 +
is called the first derived polynomial or first derivative of the
given polynomial.
The second derived polynomial or second derivative is the
first derived polynomial of the first derivative ; and so on.
The given polynomial is sometimes called the primitwe poly
nomial.
A derived equation is one whose first member is a deriva
tive of the first member of another.
1. Find the successive derivatives of x* + 5 x 1 + 3 x + 9.
Result : First, 3 x 2 + 10 x + 3.
Second, 6 x + 10.
Third, 6.
Fourth, 0.
EXAMPLES.
Find the successive derivatives of the following :
2. a; 3 5r + 6x2. 4. a x* bx % + ex Sd.
3. 2cc 2 ic7. 5. 7cc 4 13a; 2 +8xl.
EQUAL ROOTS.
512. Let the roots of the equation
x n +px n ' 1 + qx n ~ 2 + + tx 2 + ux + v = (1)
be a, b, c, Then (Art. 490), we have
x n +2? x n ~ 1 + q x n ~ 2 + = (x — a) (x — b) (x — c)
Putting x + y in place of x,
(x + y) n + p (x + y)" 1 + ... = (y + x a) (y + xb) ... (2)
By Art. 399, the coefficient of y in the first member is
nx n ~ 1 +p (n — I) x n ~ 2 + q (n — 2) x n ~* + (3)
GENERAL THEORY OF EQUATIONS. 387
which, we observe, is the first derivative of (1) ; and, as in
Art. 491, regarding x — a, x — b, as single terms, the
coefficient of y in the second member is
(x — b) (x — c) (x — d) to n — 1 factors "1
+ (x — a) (x — c) (x — d) to n — 1 factors !
+ (x — a) (x — b) (x — d) to ii — 1 factors  ^ '
+ ...... j
As (2) is identical, by Art. 413 these coefficients are equal.
Now if b = a, that is, if equation (1) has two roots equal to
a, every term of (4) will be divisible by x — a, hence (3) will
be divisible by the same factor ; therefore (Art. 486) the first
derived equation of (1) will have one root equal to a. Sim
ilarly, if c = b = a, that is, if (1) has three roots equal to a,
(3) will have two roots equal to a ; and so on. Or, in general,
If an equation has n roots equal to a, its first derived equa
tion xv ill have n — 1 roots equal to a.
513. From the principle demonstrated in Art. 512, it is
evident that to determine the existence of equal roots in an
equation we must
Find the greatest common divisor of the first member and
its first derivative. If there is no common divisor there can
be no equal roots. If there is a greatest common divisor, by
placing it equal to zero and solving the resulting equation we
shall obtain the required roots.
The number of times that each root is found in the given
equation is one more than the number of times it is found
in the equation formed from the greatest common divisor.
If the first member of the given equation be divided by the
greatest common divisor, the depressed equation will contain
the remaining roots of the original equation.
1. Find the roots of the equation
x*  14 x s + 61 x~  84 x + 36 = 0.
Here the first derivative is 4 x s — 42 x % + 122 x — 84 ; the
greatest common divisor of this and the given first member
388 ALGEBRA.
i s x 1 — 7 x + 6. Placing x 2 — 7 x + 6 == 0, we have, by the
rules of quadratics, or by factoring, x = 1 or 6. Therefore
the roots of the given equation are 1, 1, 6, and 6.
EXAMPLES.
Find all the roots of the following :
2. x sSx+13x6 = 0. 4. .T 4 6a; 2 8a;3 = 0.
3. x 3 7x 2 +16x12 = Q. 5. cc 4 24x 2 + 64z48 = 0.
514. When the equation formed from the greatest com
mon divisor is of too high a degree to be conveniently solved,
we may in certain cases compare it with its own derived
equation, and thus obtain a common divisor of a lower degree.
Of course this can only be done when the equation formed
from the greatest common divisor has equal roots.
For example, required all the roots of
x b  13 x* + 67 x 3  171 x + 216 x  108 = 0. (1)
Here the first derivative is 5 x* — 52 x 3 + 201 ,x 2 — 342 x
+ 216 ; the greatest common divisor of this and the given
first member is x 3 — 8 x 2 + 21 x — 18. We have then to solve
the equation
x 3 8x 2 +21z18 = 0. (2)
The first derivative of (2) is 3 x 2 — 16 x + 21 ; the greatest
common divisor of this and x 3 — 8 x 2 + 21 x — 18 is x — 3.
Solving x — 3 = 0, we have x = 3; hence two of the roots of
(2) are equal to 3. Dividing the first member of (2) by
(x — 3) 2 or by x 2 — 6 x + 9, the depressed equation is
x — 2 = 0, whence x = 2.
Thus the three roots of (2) are 3, 3, and 2. Hence, the five
roots of (1) are 3, 3, 3, 2, and 2.
515. If an equation has two roots equal in magnitude, but
opposite in sign, by changing the signs of the alternate terms
beginning with the second we shall obtain an equation with
these same two roots (Art. 498) ; then evidently the greatest
GENERAL THEORY OF EQUATIONS. 389
common divisor of the two first members placed equal to zero
will determine the roots.
For example, required all the roots of
x* + 3 x 3  13 x21 x + 36 = 0. (1)
Changing the signs of the alternate terms, we have
x 4  3 x 3  13 x° + 27 x + 36,
the greatest common divisor of which and the given first
member is x 2 — 9 ; solving x 2 — 9 = 0, we have x = 3 or — 3.
thus giving two of the roots of (1). Dividing the first mem
ber of (1) by x 2 — 9, we have for the depressed equation
x + 3 x  4 = 0,
whence x = 1 or — 4. Thus the roots of (1) are 3, — 3, 1,
or —4.
LIMITS OF THE ROOTS OF AN EQUATION.
516. A polynomial of the form
x n + p x n ~ * + q x n ~ 2 + J r tx 2 + ux+v
which we shall represent by X, may also be expressed thus
(Art. 490) :
(x — a) (x — b) (x — c) (x — I) Y
in which a, b, c, I are the real, unequal roots of the equa
tion X= 0, in the order of their magnitude, a being algebrai
cally the smallest; and Y the product of all the factors con
taining imaginary roots, which must always be positive, and
cannot affect the sign of X, for each pair of imaginary roots
(Art. 497) produces a positive factor.
Suppose x to commence at any value less than a, and to
assume in succession all possible values up to some quantity
greater than I. When x is less than a, each of the factors
x — a, x — b, is negative, and therefore X is either posi
tive or negative, according as the degree is even or odd.
390 ALGEBRA.
When x == a, X = 0. When x is greater than a, and less
than b, x — a becomes positive, and the sign of X changes.
Also, when the value of x is made equal to b, and then greater,
X first becomes and then changes sign; and so on, for each
real root.
When x has any value greater, than I, X must be positive ;
for all its factors are positive.
517. If two numbers, when substituted for the unknown
quantity in an equation, give results having a different sign,
at least one root lies between those numbers.
It is evident, from Art. 516, that if X has a different sign
for two values of x, some odd number of roots lies between
them.
When the numbers substituted differ by unity, it is evident
that the integral part of the root is known.
EXAMPLES.
1. What is the first figure of a root of the equation x 3 + 3 x
Here, if x = 2, the first member becomes — 2 ; and if x = 3,
the first member becomes 25 ; therefore at least one root lies
between 2 and 3. Hence 2 is the first figure of a root.
2. Find the integral parts of all the roots of the equation
x 3Gx 2 +3x+9 = 0.
3. Find the first figure of a root of the equation x 3 —2x
 50 = 0.
4. Find the first figure of a root of the equation x* — 2 ar 8
+ 3x 2 x5 = 0.
5. Find the integral part of a root of the equation 2 x 4 + x :
7x 2 llx4, = 0.
518. To find the superior limit of the positive roots of an
equation.
Let the equation be
X= x n + p x" 1 + q .r" 2 + + tx + ux + v = 0. (1)
a
GENERAL THEORY OF EQUATIONS. 301
Let r be the numerical value of the greatest negative
coefficient, and x n ~ s the highest power of x which has a nega
tive coefficient. Then the first s terms have positive coef
ficients.
Now Xwill be positive when x is positive, provided
x 11 — r x n ~ * — ) , / _,_1 — — r x — r x — r (2)
is positive ; for, since r is the numerically greatest negative
coefficient, and all terms up to the (s + l)th are positive, Xis
equal to (2) plus a, positive quantity.
We may write (2)
x n — r (x n  s + x n  s  1 + + x ° + x+l),
or (Art. 120), x*r  — — ~ . (3)
x — 1
Then AT will be positive when (3) is positive. But if x is
greater than unity, (3) is evidently greater than
x n — s + 1
Xl
Therefore X will be positive when this is positive ; or, when
(x — 1) x 11 — rx n  s + 1 is j>ositive ; or, when (x — 1) x s ~ x — r
is positive.
But (x— 1) X s1 — r is greater than (x — 1) (x — l) s_1 — r
or (x — 1)' — r ; therefore X will be positive when (x — l) s — r
is positive or equal to zero ; or, when (x — l) s = r or > r ;
or, when x — l — \Jr or > \/r; or, when x = l\\jr or >1
+ </r.
That is, when x = 1 + \J r or any greater value, X is posi
tive, which is impossible, as it must equal zero. Hence x
must be less than l + tyr; or, 1 + $r is the superior limit of
the positive roots.
519. To find the inferior limit of the negative roots of an
equation.
By changing the signs of the alternate terms beginning
with the second, we shall obtain an equation having the
same roots with contrary signs (Art. 408).
392 ALGEBRA.
Then evidently the superior limit of the positive roots of
the transformed equation, obtained as in Art. 518, will by a
change of sign become the inferior limit of the negative roots
of the given equation.
Note. In applying the principles of the preceding articles to determine
the limits of the roots of an equation, the absolute term must be taken as
the coefficient of x'K
520. 1. Find the superior limit of the positive roots of
x 4 + 4 x 3  19 x  46 x + 120 = 0.
Here, r = 46, and n — s = 2 ; or, as n = 4, s — 2. Then by
Art. 518, the required limit is 1 + y/46, or 8 in whole num
bers.
2. Find the inferior limit of the negative roots of
a 3 a; 2 14.x +24 = 0. (1)
Changing the signs of the alternate terms beginning with
the second, we have
x 3 + x 2 Ux2A = 0. (2)
Here r = 24, and n — s = 1, or s = 2. Then the superior
limit of the positive roots of (2) is 1 + \/24 ; therefore the
inferior limit of the negative roots of (1) is — (1 + v/24).
EXAMPLES.
Find the superior limits of the positive roots of the follow
ing:
3. x * + 2x 3 13.r' 1 Ux + 24:=Q. 4. a: 4 15.T 2 +10,r+24=0.
Find the inferior limits of the negative roots of the fellow
s'
mg
5. x*2x°5x + (j = 0. 6. x x 5x 5 +ox + 5x + 6 = 0.
STURM'S THEOREM.
521. To determine the number and situation of the real
roots of an equation.
GENERAL THEORY OF EQUATIONS. 393
A perfect solution of this difficult problem was first obtained
by Sturm, in 1829. As the theorem determines the number
of real roots, the number of imaginary roots also becomes
known (Art. 487).
522. Let X denote the first member of
X n +21X n  1 + qx n ~ 2 + + tX 2 + UX+ V = 0,
from which the equal roots have been removed (Art. 512).
Let X x denote the first derivative of X (Art. 511).
Divide AT by X 1} and we shall obtain a quotient Q l} with a
remainder of a lower degree than X v Denote this remainder,
with its signs changed, by X 2 , divide X x by X 2 , and so on ;
the operation being the same as in finding the greatest com
mon divisor, except that the signs of every remainder must be
changed, while no other change of signs is admissible. As
the equation X — has been freed from equal roots, there can
be no common divisor of Zand J 1; and the last remainder,
X„ , will be independent of x.
The successive operations may be represented by the fol
lowing equations :
X = X x Q v  X 2 (1)
X I = X 2 Q 2 X 3 (2)
X^X,Q,X 4 (3)
X n _ 2 — <*«  1 V«  1 X n
The expressions X, X 1} X 2 , X H are called Sturm's
Functions.
STURM'S THEOREM.
523. If any tiro numbers, a and b, be substituted/or x in
Sturm's Functions, and the signs noted, the difference between
the number of variations in the first case and that in the
second is equal to the number of real roots of the given equa
tion lying between a and b.
394 ALGEBRA.
The demonstration of Sturm's Theorem depends upon the
following principles :
(A). Two consecutive functions cannot both become for
the same value of x.
For, if A\ = and X = 0, then by (2), Art. 522, X 3 = ;
and if X 2 = and X 3 = 0, by (3), X 4 = 0; and so on, till
X n = 0. But as X n is independent of x, it cannot become
for any value of x. Hence no two consecutive functions can
become zero for the same value of x.
(B). If any function, except X and X n , becomes for a
■particular value of x, the two adjacent functions must have
opjjosite signs.
For, if X, = 0, we have by (2), Art. 522, X 1 = — X s ; that
is, X 1 and X s must have opposite signs, for by (A) neither can
be equal to zero.
(C). When any function, except X and X n , changes its sign
for different values of x, the number of variations is not
affected.
No change of sign can take place in any one of Sturm's
Functions except when x passes through a value which re
duces that function to zero.
Now, let c be a root of the equation X 2 = 0; d and e quan
tities respectively a little less and a little greater than c, so
taken that no root of X x = or of X s = is comprised be
tween them. Then, as x changes from d to e, no change of
sign takes place in A^ or X :i , while X 2 reduces to zero and
may change sign. And as by (B), when X 2 = 0, X l and X z
have opposite signs, the only effect of a change in the sign of
X 2 is that what was originally a permanence and a variation
is now a variation and a permanence ; that is, the permanence
and variation exchange places. Hence a change in the sign
of X., does not affect the number of variations.
As X n is independent of x, it can never change sign for any
value of x. Therefore a change in the number of variations
GENERAL THEORY OF EQUATIONS. 395
can be caused only by a change in the sign of the given
function X.
(D). When the function X changes its sign for successive
increasing values of x, the number of variations is diminished
by one.
Let m be a root of the equation X = ; m — y and m + y
quantities respectively a little less and a little greater than
7>i, so taken that no root of X 1 = Q is comprised between them.
Then, as x changes from m — y to m + y, no change of sign
takes place in X r , while X reduces to zero and changes sign.
Putting in + y in place of x in X, we have
O + y) n + P (m + y) n ~ l + + u (m + y) + v.
B/eveloping the terms by the Binomial Theorem, and col
lecting terms containing like powers of y, we have
m n + p m n ~ 1 + + um + v
+ y \_n m" _ l + p (n — 1) ni n ~~ 2 + + u\
+ terms containing y 2 , y s , y n .
Representing the coefficient of y, which we observe is the
value of X 1 when x is put equal to m, by A ; the coefficient of
y 1 by B; and so on, we have
m n +pm n  1 + + um + v + Ay + By+ + Ky n . (1)
But as x = m reduces X to 0, we have
m n + p m n ~ 1 + + u m + v = 0.
Hence (1) may be written
Ay + Bf+ + Ky\ (2)
Now y may be taken so small that the sign of (2) will be
the same as the sign of its first term. That is, when a? is a
little greater than m, the sign of X is the same as the sign
of X l .
Similarly, by substituting in — y for x in X, we shall arrive
at the expression
Ay+By 2 Cy 3 + ,
396 ALGEBRA.
where as before y may be taken so small that the sign of the
whole expression will be the same as that of its first term.
That is. when a; is a little less than m, the sign of X is the
reverse of the sign of X x .
Thus we see that as x changes from m — y to m + y, the
signs of X and A^ are different before x equals m, and alike
afterwards. Hence, when A changes its sign a variation is
changed into a permanence, or the number of variations is
diminished by one.
We may now prove Sturm's Theorem ; for as x changes
from a to b, supposing a less than b, a variation is changed to
a permanence each time that X reduces to and changes sign,
and only then, for no change of sign in any of the other
functions can affect the number of variations. And as X
reduces to zero only when x is equal to some root of the
equation X — 0, it follows that the number of variations lost
in passing from a to l> is equal to the number of real roots of
the equation X = comprised between a and b.
524. When — oo and + go are substituted for x, or when
the superior limit of the positive roots and the inferior
limit of the negative roots are substituted for x, the whole
number of real roots of the equation A^ = becomes known.
The substitution of — go and will give the whole number
of negative roots, and the substitution of + go and will give
the whole number of positive roots. If the roots are all real,
Descartes' Kule (Art. 507) will effect the same object.
The substitution of various numbers for x will show be
tween what numbers the roots lie, or fix the limits of the
roots.
525. A and A, must change signs alternately, as they are
always unlike in sign just before X changes sign (Art. f>L\' > > i
(D)). Hence, when the roots of X = and of Aq = are all
real, each root of Aq = must be intermediate in value be
tween two roots of X= 0. For this reason the first derived
equation is often called the limiting or separating equa
tion.
GENERAL THEORY OF EQUATIONS. 397
526. In the process of finding X 2 , X 3} etc., any positive
numerical factors may be omitted or introduced at pleasure, as
the sign of the result is not affected thereby. In this way
fractions may be avoided.
In substituting — go and + go, the first term of each func
tion determines the sign, for in any expression, as
ax n + bx n ~ 1 + + /,•,
where x may be made as great as we please, it may be taken
so great that the sign of the whole expression will be the same
as that of its first term.
527. 1. Determine the number and situation of the real
roots of the equation
x 3 — 4 x 2 — x + 4 = 0.
Here, the first derivative, X 1 = 3 x 2 — 8 x — 1. Multiply
ing x 3 — 4 x' 1 — x + 4 by 3 so as to make its first term divisi
ble by 3 x 2 ,
3x 2 8xl)3x 3 12x 2  3 a: + 12 (a;
3 a; 3  Sx 2
x
 4r 2 a; + 12
3
 6 a; 2  3 x + 18 (2
 6.r + 16a + 2
19 a: + 16 .. X = 19 x  16.
3 x 2  8 x  1
19
19 x  16 ) 57 x 2  152 x  19(3x
57 x 2 — 48 x
 104 x  19
19
1976 a; 361 (104
1976 x + 1664
2025 .. X = 2025.
398 ALGEBRA.
Thus we have, X = x z — 4 x 2 — x + 4 ; X 2 = 19 cc — 10.
Xj == 3 x  8 x  1 ; Z 3 = 2025.
The last step of the division may he omitted, for we only
wish the sign of X 3 , and that may he seen by inspection when
— 104 x — 19 is obtained.
We first substitute — go for x in each function, and obtain
three variations of sign ; similarly + go gives no variation ;
hence the three roots are all real. Substituting 0, we have
two variations ; comparing this with the former results, we
see that one root is negative and the other two are positive.
The same result could have been obtained by Descartes' Rule,
as all the roots are real. We now substitute various numbers
to determine the limits of the roots.
The table presents the results in a connected form :
X
*1
^2
*3
When
X = — 00,
—
+
—
+
3 variations.
Li
x = — 2,
—
+
—
+
3 variations.
a
x = — 1,
+
—
+
it
x = 0,
+
—
—
+
2 variations.
n
X = 1,
—
+
+
a
x = 2,
—
—
+
+
1 variation.
a
% == O}
—
+
+
+
1 variation.
u
x = 4,
+
+
+
a
OC — Oj
+
+
+
+
no variation.
a
X = GO,
+
+
+
+
no variation.
Then by Sturm's Theorem we know that there is one root
between —2 and 0, one between and 2, and one bet ween 3
and 5. In fact, as X == when x = — 1, 1, and 4, these are
the three roots of the equation.
2. Determine the number and situation of the real roots of
X
x 4  3 x s + 3 x — 3 x +  =
Note. In substituting the various numbers to determine the situation
of the roots, it is best to work from o in cither direction, stopping when
the number of variations is the same as has been previously found for
+ oo or — oo , as the case may be.
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 399
Here we find X, = 4x 3  9x 2 + 6x  3 ; X a = — <d2x + 129 ;
X = 3 x 2 + 18 x  31 ; X i =  1163.
Substituting + co for x, we obtain one variation ; similarly,
gives three variations, and — oo gives three variations.
Hence there are only two real roots, both of which are posi
tive. We then substitute values of x from upwards, giving
the following results :
X
^
x 2
X s X 4
When x = 0,
+
—
—
+ —
3 variations.
■ " x = l,
+
—
—
+ 
3 variations.
" x = 2,
+
+
+
— —
1 variation.
" x = co,
+
+
+
— —
1 variation.
Hence there are two roots between 1 and 2 ; and as the
equation has four roots, there must be two imaginary roots.
EXAMPLES.
Determine the number and situation of the real roots of the
following equations :
3. x*x 2 2x + l = Q. 6. x*2x 9 5x 2 + 10a;3 = 0.
4. g* — 7x+7 = 0. 7. 2x 4 3* 3 + 17x 2 3x + 15 = 0.
5. X s — 2 a 5 = 0. 8. x* 4 X s 3 x + 27 = 0.
XLIIL — SOLUTION OF HIGHER NUMERICAL
EQUATIONS.
528. The real roots of the higher numerical equations in
general can only be obtained by tentative methods, or by
methods which involve approximation. Cubic and biquadratic
equations may be considered as included in the class of higher
400 ALGEBEA.
equations ; for their general solutions are complicated, and
only of limited application. No general solution of an equa
tion of a degree higher than the fourth can he ohtained.
COMMENSURABLE ROOTS.
529. A commensurable root is one which can be exactly
expressed as an integer or fraction without using irrational
quantities.
An incommensurable root is one which can only he ex
pressed approximately by means of a decimal fraction.
530. Any equation containing fractional coefficients may
be transformed into another whose coefficients are entire, that
of the first term being unity (Art. 500), and such an equation
cannot have a root equal to a rational fraction (Art. 495) ;
hence, to find all commensurable roots, we have only to find
all integral roots.
531. As every rational root of an equation in its general
form is a divisor of the last term (Art. 493), to find the com
mensurable roots we have only to ascertain by trial what in
tegral divisors of the absolute term are roots of the equation.
The trial may be made by substituting each divisor, both
with the positive and the negative sign, in the equation ; or
by dividing the first member of the equation by the unknown
quantity minus the supposed root (Art. 486). In substi
tuting very small numbers, such as ± 1, the former method
may be most convenient ; but when an actual root lias once
been used, the latter method will give at once the depressed
equation, which may be used in obtaining the other roots.
532. When the number of divisors of the last term is
large, this process of successive trials becomes tedious, and a
better method, known as the Method of Divisors, may be
adopted.
If a is a root of the equation
x 4 +px i +qx 2 + tx + u = 0,
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 401
then a 4 + p a s + q a + t a + u= 0.
Transposing and dividing by a,
= — t — qa—pa 2 — a 3 , (1)
u
whence we see that  must be an integer.
a
Equation (1) may be written
— h t = — q a —p a — a*,
cv
U
a
a
u
Denoting  + t by t', and dividing by a,
— — q —p a — a 2 ,
t'
whence  must be an integer.
a
Proceeding in this way, we see that if a is a root of the
equation, — \ t or t', — \ q or q', and \ p or p 1 must be in
Ct Ct Lt
p'
tegers, and 11 must equal zero.
Hence the following
RULE.
Divide the absolute term of the equation by one of its inte
gral divisors, ami to the quotient add the coefficient of x.
Divide this sum by the same divisor, and, if the quotient is
an integer, add, to it the coefficient ofx 2 .
Proceed in the same manner with each coefficient in regular
order, and, if the divisor is a root of the equation, each
quotient will be entire, and the last quotient added to the
coefficient of the highest poicer ofx will equal 0.
Equal roots, if any, should be removed before applying the
rule ; and the labor may often be diminished by obtaining the
superior limit to the positive and inferior limit to the nega
402 ALGEBRA.
tive roots of the equation, for no number need be tried which
does not fall between these limits.
1. Find the roots of the equation
x s  6 x 2 + 27 x  38 = 0.
By Descartes' Rule, we see that the equation has no nega
tive root; and the only positive divisors of 38 are 1, 2, 19,
and 38. By substitution we see that 1 is not a root of the
equation.
Dividing the first member by x — 2, we obtain x 2 — 4 x + 19
as a quotient. Hence 2 is a root, and the depressed equation
is x 2 — 4 x + 19 = 0, from which we obtain
x =
4±y/1676
~~2~
as the remaining roots. Hence,
2±v/15
x = 2, or 2 ± y/ — 15, Ans.
2. Find the roots of the equation
8 x*  4 x s  14 x 2 + x + 3 = 0.
We may write the equation
A 9C i QC 0C O 
*2^r + § + 8= 
Proceeding as in Art. 500, we see that the multiplier 2 will
remove the fractional coefficients. We then have the equation
x 4 — x s — 7 x 2 + x + 6 = 0, (1)
whose roots are twice those of the given equation (Art. 499).
The divisors of G are ±1, ±2, ±3, and ± 6.
By putting x equal to + 1 and — 1 in (1), it is readily seen
that both are roots of the equation, and the other roots can be
found from the depressed equation. But all of the rational
roots may be obtained by the rule.
6,
3,
2,
1,
1,
2,
3,6
1,
2,
3,
6,
6,
3,
2, 1
2,
3,
4,
("7
5,
2,
1,
1,
2,
7,
5,
1,
6,
$
o,
9
6,
7
2,
o,
2,
3
3,
1,
1,
2
1,
1,
1,
1
o,
o,
o,
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 403
It is customary to abridge the work as follows :
Divisors,
1st Quotients,
Adding 1,
2d Quotients,
Adding — 7,
3d Quotients,
Adding — 1,
4th Quotients,
Adding 1,
As 6, 2, — 3, and — 6 give fractional quotients at different
stages of the operation, they cannot be roots of the given
equation, and are rejected. 3, 1, — 1, and — 2 give entire
quotients, and in each case the last quotient added to the
coefficient of x 4 gives zero ; hence they are the four roots of
3 1 1
equation (1), and ^, ^, — ^, and — 1 are the four roots of the
given equation.
EXAMPLES.
Find all the commensurable roots of the following equa
tions, and the remaining roots when possible by methods
already given :
3. x s +Qx 2 +llx + Q = 0. 8. x s 7x 2 + 36 = 0.
4. x s + 3x 2 4;x12 = 0. 9. x 3 6x 2 + 10a; — 8 = 0.
5. a 4 4.r 3 8x + 32 = 0. 10. x 3  6 x 2 + 11 x 6 = 0.
6. 4a; 8 16sc 2 9a; + 36 = 0. 11, 2x 3 3x 2 +16x2i = 0.
7. x 3 3x 2 + x + 2 = 0. 12. x 5  2 x 3 16 = 0.
13. x i 9x 3 + 23x 2 20x + 15 = 0.
14. x 4 + x 3  29 x 2  9 x + 180 = 0.
404 ALGEBRA.
RECURRING OR RECIPROCAL EQUATIONS.
533. A Recurring Equation is one in which the coef
ficients of any two terms equally distant from the extremes
of the first member are equal.
The equal coefficients may have the same sign, or opposite
signs; hut a part cannot have the same sign, and a part
opposite signs, in the same equation. Also, if the degree he
even, and the equal coefficients have opposite signs, the middle
term must be wanting. Thus,
a 4  5 x z + 6 x 2  5 x + 1 = 0,
5 x 5  51 x* + 160 x 3  160 x 2 + 51 x  5 = 0,
x 6 — x 5 + x* — x 2 + x — 1 = 0,
are recurring equations.
534. If any quantity is a root of a recurring equation,
the reciprocal of that quantity is also a root of the same
equation.
Let x n +px n ~ 1 + qx n ~ 2 + ...± (... + </x 2 +i*z + l)=0 (1)
be the equation. Substitute  for x ; then
V
y" y'" 1 y n ~ v v v >
Multiplying each term by y n ,
(i+py+.qf.+,..)±(...+ qy n *+py n  1 +y n )=Q ( 2 )
Now, (1) and (2) take precisely the same form on changing
the ± sign to the first parenthesis in equation (2), and hence
they must have the same roots. Now, if a is a root of (1), as
1 1
v —  must be a root of (2) ; but, as (1) and (2) have the
x a
same roots,  must also be a root of (1). In like manner, if
a
b is a root of (1), y is also a root of (1).
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 405
On account of the property just demonstrated, recurring
equations are also called reciprocal equations ; the former term
relating to their coefficients, and the latter to their roots.
535. One root of a recurring equation of an odd degree is
— 1 when the equal coefficients have the same sign, and +1
when they have opposite signs.
A recurring equation of an odd degree, as
x *n + l +px 2m + qx 2ml + _ ± ( + q tf + p x + ^ _ Q (3)
has an even number of terms, and may he written in one of
the following forms,
(x 2M + 1 + 1) +p (x 2m + x) + q (x 2 "' 1 + x") + =0,
( X 2m + 1 — 1) +p (x 2m — x) + q (X lm  1 X) + ...... = 0.
If — 1 he substituted for x in the first form, or + 1 in the
second, the first member will become ; hence, — 1 is a root
of the first and + 1 a root of the second.
If equation (3) be divided by x ± 1, both forms will reduce
to the following form,
X 2m +px 2 " 1  1 + qx 2m ~ 2 + + qx 2 +px + l = 0, (4)
a recurring equation of an even degree in which the equal
coefficients have the same sign. Hence, a recurring equation
of an odd degree may always be depressed to one of an even
degree.
536. Two roots of a recurring equation of an even degree
are + 1 and — 1 when the equal coefficients have opposite signs.
Let
x *>» + pX 2m  1 + qX 2m  2 + ( + qx 2 +px+ 1)=0
be such an equation. As the middle term must be wanting
(Art. 533), the equation may be written in the form
^™l)+ F (^»" ! l) + ?x 2 (f'» 4 l)h = (5)
which is divisible by both x — 1 and x + 1, or by x' 2 — 1 (Art.
120). Hence, both + 1 and — 1 are roots of the equation.
406 ALGEBRA.
If equation (5) be divided by x 1 — 1, it will be depressed
two degrees, and become a recurring equation of an even
degree, in which the equal coefficients have the same sign
(Art. 120). Hence, every recurring equation may be de
pressed to the form of equation (4), Art. 535.
537. Every recurring equation of an even degree, whose
equal coefficients have the same sign, may be reduced to an
equation of half that degree.
Let
a? m +qi x"' 1 + q ar m ~ 2 + + q x~ +p x + 1 =
be such an equation. Dividing it by x m , we may write it
(^ + ^) + ^(^" , + ^)+?(^ 2 + ^r 2 ) + =0 (6)
the middle term if present becoming a known quantity.
Put x +  = y
Then, a 2 + = = y 2  2
x
x3+ x^ =y3 ~ s ( x+ x) =i/3 ~ s y
x m + — = y m — m y m ~ 2 +
Substituting these values in (6), we have an equation of the
form
y" l +2hy m  l + qiy m * + = 0.
After this equation is solved, we can immediately find x
from the equation x +  = ?/.
x
538. It thus appears that any recurring equation of the
(2 m + l)th degree, one of the (2 m + 2)th degree whose equal
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 407
coefficients have opposite signs, and one of the 2 with degree
whose equal coefficients have the same sign, may each he
reduced to an equation of the with degree.
EXAMPLES.
1. Given x 4 — 5 x s + 6 x 2 — 5 x + 1 = 0, to find x.
Dividing by x 2 , far \ — j j — 5 f x \ — j + 6 = 0.
S instituting y for x \ — , and y 2 — 2 for x 2 \ —  9 we have
x x l
w 2 25w+6 = 0.
Whence, y = 4 or 1.
If y = 4, x +  = 4, or x 2 — 4 x = — 1 ;
Whence, x = 2 ± ^3.
If y = 1, x +  = 1, or a; 2 — x = — 1 ;
Whence, x = T
Note. That 2 — y/3 and are reciprocals of 2 + y/3 and  — ^
1 2
may easily be shown by reducing — — — and —  — ;=. to equivalent frac
2 + ^3 1 + V— 3
tions with rational denominators (Art. 279).
Solve the following equations :
2. a 5  11 x l + 17 x 3 + 17 a 2  11 x + 1 = 0.
3. a; 5 + 2 x*  3 a; 3  3 x 2 + 2 x + 1 = 0.
4. x G — x 5 + x 4 — x 2 + x — 1 = 0.
5. a; 3 + px 2 +px + 1 = 0.
6. 6 x 4 + 5 a. 3  38 x 2 + 5 x + 6 = 0.
7. 5 x 5  51 a, 4 + 160 a; 3  160 a 2 + 51 x  5 = 0.
408 ALGEBRA.
8. x 4 + 5 x' + 5 x + 1 — 0.
9. x 5 =  1, or x> + 1 = 0. (See Art. 332.)
10. x 5 32 = 0. (Let cc = 2 ?/.)
CARDAN'S METHOD FOR THE SOLUTION OF CUBIC
EQUATIONS.
539. In order to solve a cubic equation by Cardan's
method, it must first be transformed, if necessary, into another
cubic equation 1 in which the square of the unknown quantity
shall be wanting.
By Art. 505, this may be done by substituting for x, y
minus the coefficient of x 2 divided by 3.
540. If the first power of the unknown quantity be want
ing in the given equation, we may obtain the result by a
simpler method, as follows :
Let x 3 + a x 2 + c = be such an equation.
Substituting  for x, we have
V
1 a
— + — + c = 0, or c y % + a y + 1 = 0.
541. To solve a cubic equation in the form x 3 + p x + q = 0.
P
Put x = z — =, and the equation becomes
Sz
V 2 V 3 7) 2
or, z 3  £—. + q = 0; or, 27 z* + 27 q z 3 p 3 = 0.
2( z 3
This is an equation in the quadratic form, and may be
solved by the method of Art. 313 ; and after z is known, x
P
may be found directly from the equation x = « — —.
o z
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 409
We have then for solving cubic equations the following
RULE.
If necessary, transform the equation into another cubic
equation in which the square of the unknown quantity shall
be wanting (Arts. 539 and 540).
If y be the unknown quantity in the resulting equation,
substitute for it z minus the coefficient of y divided by 3z.
EXAMPLES.
1. Solve the equation x 3 — 9 x + 28 = 0.
3
Substituting z \ — for x,
27 97 27
z 3 + 9z + — +^9z  — + 28 = 0,
z z 6 z
27
or, « 3 + tf + 28 = 0; or, * c + 28s 3 = 27.
Solving by quadratics, z 3 = — 1 or — 27.
Whence, z = — 1 or — 3.
Uz = 1, x = z +  = l3 = ±.
z
If z = 3, x =  3  1 =  4.
Hence, one root of the equation is — 4. Dividing the first
member of the given equation by x + 4, we obtain as the de
pressed equation,
x 2 — 4 x + 7 = 0.
Whence, x — 2 ± y/— 3, the remaining roots.
2. Solve the equation x s — 24 x 2 — 24 x — 25 = 0.
Putting x = y + 8 (Art. 539), we obtain
2/ 3 + 24^+192 Z / + 51224y 2 384?/153624y19225=0,
or, y 8  216 y — 1241 = 0.
410 ALGEBRA.
72
Putting y = z H — , we have
. _,_ 15552 373248 ._ 15552 , OM _
z z + 216s h 1 5 216 « 1241 = 0,
z z* z
or, z 3 + 3 ' 3 ^ 48  1241 = ; or, s c  1241 z 3 + 373248 = 0.
Whence, s s = 729 or 512, and 2 = 9 or 8.
72 72
Therefore, y = 9 +  r or 8 + ^ = 17, and sc = y + 8 = 25.
9 o
Hence, one root of the equation is 25. Dividing the first
member of the given equation by x — 25, we have as the
depressed equation
x 2 + x + 1 = 0:
 1 ± \J 3 ,
Whence, x =  , the remaining roots.
ii
Solve the following equations :
3. x 3  6x + 9 = 0. 6. x 3 + 9 x 2  21 x + 11 = 0.
4. x 3  6 x 2 + 57 x 196=0. 7. x 3 2 x 2 + 2x 1 = 0.
5. :c 3 4a; 2 3x + 18 = 0. 8. x 3 ±x 2 + 4a; 3 = 0.
9. a 3 3x 2 + 4 = 0.
10. Obtain one root of the equation x 3 + 6 x — 2 = 0.
542. In the cubic equation x 3 + px+q = 0, when p is
— p 3 o 2
negative, and f, > =y , Cardan's method involves imaginary
expressions ; but it may be shown in that case that the three
roots of the equation are then real and unequal.
Thus, in solving the equation x 3 — 6 x + 4 = 0.
2
Substituting z +  for x, we have
z
z 3 + 6 2 H \ j — Gz f 4 = 0,
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 411
r, g« + J* + 4 = ; or, z r > + 4 z* + 8 = 0.
Whence, z z = 2 ± y/ 4, or 2 ± 2 y/ 1,
or, ft = y/_ 2 + 2y/ 1 or ^22^1.
It may be proved by trial that 1 + y/— 1 is the cube root of
_ 2 + 2 yC3, and 1  y/^1 of  2  2 yCl. Hence,
= l + y/l or ly/l.
If z = 1 + y/ 1,
X=Z+=1+ y/— 1 +
2 _2y/l + 2
i + ^/=3 _ ' 1 + y'i
_ o
Hence, one root of the equation is 2. Dividing the first
member of the given equation by x — 2, we have as the de
pressed equation
x 2 + 2 x  2 = 0.
Whence, £C = — 1 ± y^, the remaining roots.
543 We have no general rule for the extraction of the
cube root of a binomial surd ; so that in examples like that
in the preceding article, unless the value of z can be obtained
by inspection, it is impossible to find the real values of x by
Cardan's method. . In this case, the real values of x can
always be found by a method involving Trigonometry.
BIQUADRATIC EQUATIONS.
544. General solutions of biquadratic equations have been
obtained by Descartes, Simpson, Euler, and others. Some of
them require the second term of the equation to be removed,
while others do not. All of them depend upon the solution of
a cubic equation by Cardan's method, and will of course fail
when that fails (Art. 542). They are practically of little
value, especially as numerical equations of all degrees can be
readily solved by methods of approximation.
412 ALGEBRA.
INCOMMENSURABLE ROOTS.
545. If a higher numerical equation is found to contain
no commensurable roots, or if, after removing the commen
surable roots, the depressed equation is still of a higher
decree, the irrational or incommensurable roots must next be
sought. The integral parts of these roots may be found by
Sturm's Theorem or by Art. 517, and the decimal parts by
any one of the three following methods of approximation.
HORNER'S METHOD.
546. Suppose a root of the equation
x" +j>x"~ 1 + q x n ~ 2 + + tx' 2 + ux + v = (1)
is found to lie between a and a + 1. Transform the equation
into another whose roots shall be less by a (Art. 502), and we
shall have an equation in the form
y n +p'V n ~ 1 + Q'y n ~ 2 + + t'y° + u'y + v' = 0, (2)
one of whose roots is less than 1. If that root is found to lie
between the decimal fractions a' tenths and a' + 1 tenths,
transform equation (2) into another whose roots shall be less
by a' tenths, and we shall have an equation in the form
s»+_p"2" 1 + q"z n ~ 2 + ...... + t"z 2 + u"z + v" = (3)
one of whose roots is less than .1. If that root is found to
lie between the decimal fractions a" hundredths and a" + 1
hundredths, transform equation (3) into another whose roots
shall be less by a" hundredths ; and so on.
Thus we obtain
x = a + a' + a" +
to any desired degree of accuracy.
As y and z in equations (2) and (3) are fractional, their
higher powers are comparatively small ; hence approximate
values of y and z may be found by considering the last two
terms only, from which we have
v' .. v"
V = 77 and * ^ ~ TJ, '
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 413
Thus approximate values of a', a", may be found in
this way, and with greater accuracy the smaller they become.
Hence a positive incommensurable root of the equation may
be found by the following
RULE.
Find by Sturm's Theorem the initial part of 'the root, and
transform the given, equation into one whose roots are less by
th is initial part.
Divide the absolute term of the transformed equation by
the coefficient of the first power of the unknown quantity for
the next figure of the root.
Transform this last equation into another whose roots are
less by the figure of the root last found, divide as before for
the next figure of the root ; and so on.
547. A negative root may be found by changing the
signs of the alternate terms of the equation beginning with
the second, and finding the corresponding positive root of the
transformed equation (Art. 498). This hy a change of sign
becomes therequired negative root.
548. In obtaining the approximate value of any one of
the quantities a 1 , a", by tbe rule, we are liable to get too
great a result ; a similar case occurs in extracting the square
or cube root of a number. We may discover such an error
by observing the signs of the last two terms of the next
transformed equation ; for, as the figures of the root as ob
tained in succession are to be added, it follows that a', a",
must be positive quantities, so that the last two terms of the
transformed equation must be of opposite sign. We then
diminish the approximate value until a result is found which
satisfies this condition.
549. If in any transformed equation the coefficient of the
first power of the unknown quantity should be zero, the next
figure of the root may be obtained by dividing the absolute
term by the coefficient of the square of the unknown quantity,
and taking the square root of the result.
414 ALGEBRA.
For, if in equation (2), Art. 546, u' = 0, we have, approxi
mately,
t' y 2 + v' = 0, whence y = \/
We proceed in a similar manner if any numher of the
coefficients immediately preceding the absolute term reduce
to zero.
550. 1. Solve the equation x 3 — 3 x 2 — 2 x + 5 — 0.
By Sturm's Theorem, the equation has three real roots ;
one between 3 and 4, another between 1 and 2, the third
between — 1 and — 2.
To find the first root, we transform the equation into
another whose roots are less by 3, which by Art. 503 is
effected as follows :
Dividing x 3 — 3 x 2 — 2 x + 5 by x — 3, we have x 2 — 2 as a
quotient and — 1 as a remainder. Dividing x 2 — 2 by x — 3,
we have x + 3 as a quotient and 7 as a remainder. Dividing
x + 3 by x — 3, we have 1 as a quotient and 6 as a remain
der. Hence the transformed equation is
x 3 + 6 x 2 + 7 x  1 = 0,
whose roots are less by 3 than those of the given equation.
Note. The operations of division in Horner's Method are usually per
formed by a method known as Synthetic Division. For example, let it be
required to divide ofi  19 x + 30 by x  2.
* 3 ±0.r 2 19a; + 30
T 3 _ O r 2
x 2
a; 2 + 2x15
2 a; 2
2.C 2  ix
15a;
15.Z + 30
The first term of each partial product may be omitted, as it is merely a
repetition of the term immediately above. Also the remaining term of
each partial product may be added to the corresponding term of the divi
dend, provided we change the sign of the second term of the divisor before
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 415
multiplying. Also the powers of x may be omitted, as we need only
consider the coefficients in order to obtain the remainder.
The work now stands
l±019 + 30l+2
+ 2  1 + 215
+ 2
+ 4
15
30
As the first term of the divisor is 1, it is usually omitted, and the first
terms of the dividends constitute the quotient. Raising the oblique
columns we have the following concise form :
Dividend, l±019 + 30+2
Partial Products, +2+ 430
Quotient, 1 + 215,+ Remainder.
Here we use only the second term of the divisor with its sign changed ;
each term of the quotient is the sum of the terms in the vertical column
under which it stands, and each term of the second line is obtained by
multiplying the preceding term of the quotient by the divisor as written.
By the method of Synthetic Division, the work of trans
forming the given equation into one whose roots are less by
3 stands as follows :
1 
3
—
2
+ 5 +3
+
3
 6
—
2
— 1, 1st Remainder
+
3
+
9
+
3
+
7,
2d Remainder.
+
3
f 6, 3d Remainder.
Thus the transformed equation is, as before,
x 3 + 6 x 2 + 7 x  1 = 0. (1)
Dividing 1 by 7 we obtain .1 as the next figure of the root,
and we proceed to transform equation (1) into another whose
roots shall be less by .1.
416
ALGEBRA.
6
7
1
.1
.61
.761
6.1
7.61
.239
.1
.62
6.2
8.23
.1
6.3
Thus the transformed equation is
x s + 6.3 x 2 + 8.23 x  .239 = 0,
whence by dividing .239 by 8.23 we obtain .02 for the next root
figure ; and so on. Thus the first root is, approximately, 3.12.
Similarly, the second root may be shown to be 1.201 ap
proximately.
By Art. 547, the third root is the positive root of the
equation x 3 + 3 x~ — 2 x — 5 — with its sign changed. The
successive transformations are usually written in connection
as in the following form, where the coefficients of the different
transformed equations are indicated by (1), (2), (3),
The work may also be contracted by dropping such decimal
figures from the right of each column as are not needed for
the required degree of accuracy.
1
3
_ 9
4
5  1.33
1
2
4
2
(1)173
1
5
2.667
5
(1)7
(2) .333
1
1.89
(1)6
8.89
.3
1.98
6.3
(2) 10.87
.3
6.6
.3
(2) 6.9
Hence, the third root is — 1.33 approximately.
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 417
EXAMPLES.
Find the real roots of the following equations :
2. a r 3 _2x5 = 0. 5. x 3  17 x 2 + 54 x 350 = 0.
3. a* + x 3  500 = 0. q a; 4 4 a 3 3 a: + 27 = 0.
4. x 3  7 a; +7 = 0. 7. x 4  12 x 1 + 12 x 3 = 0.
APPROXIMATION BY DOUBLE POSITION.
551. Find two numbers, a and b, the one greater and the
other less than a root of the equation (Arts. 517 or 521), and
suppose a to he nearer the root than b. Substitute them
separately for x in the given equation, and let A and B repre
sent the values of the first member thus obtained. If a and b
were the true roots, A and B would each be ; hence the
latter may be considered as the errors which result from sub
stituting a and b for x. Although not strictly correct, yet,
for the purpose of approximation, we may assume that
A : B = x — a :x — b
Whence (Art. 348), A — B : A = b — a : x — a
or (Art. 345), A — B : b — a = A:x — a (1)
A(ba)
and, x — a = — —
A — B
A(ba)
or, x = a + A _ B .
From (1), we see that, approximately,
As the difference of the errors is to the difference of the two
assumed numbers, so is either error to the correction of its
assumed number.
Adding this correction when its assumed number is too
small, or subtracting when too large, we obtain a nearer
approximation to the true root. This result and another
418 ALGEBRA.
assumed number may now be used as new values of a and b,
for obtaining a still nearer approximation ; and so on.
It is best to employ two assumed quantities tbat sball differ
from each other only by unity in the last figure on the right.
It is also best to use the smaller error.
This method of approximation has the advantage of being
applicable to equations in any form. It may, therefore, be
applied to radical and exponential equations, and others not
reduced to the general form (Art. 480).
EXAMPLES.
1. Find a root of the equation x z + x + x — 100 = 0.
When 4 and 5 are substituted for x in the equation, the
results are — 16 and + 55, respectively; hence a = 4, b = 5,
A — — 16, and B = 55. According to the formula,' the first
approximation gives
_ 16 (5  4) , 16 , _
* = 4+ 1655  4 + 7l = 42+ 
As the true root is greater than 4.2, we now assume 4.2
and 4.3 as a and b. Substituting these values for x in the
given equation, we obtain — 4.072 and + 2.297 ; therefore 4.3
is nearer the true root than 4.2.
„ , Q 2.297(4.34.2) .2297
Hence, x = 4.3 ^ + 4 _ = 4.3  ^
4.3 .036 = 4.264.
Substituting 4.264 and 4.265 for x, and stating the result
in the form of a proportion, we have
.0276 + .0365 : .001 = .0270 : correction of 4.264.
Whence the correction = .00043+.
Hence, x = 4.264 + .00043 = 4.26443+, Am.
Find one root of each of the following equations :
SOLUTION OF HIGHER NUMERICAL EQUATIONS. 419
2. x 3  2 x 50 = 0. 4. a; 3 + 8r+ Gx — 75.9 = 0.
3. * 3 +10x 2 +5a;260 = 0. 5. x 3 +^^ T = 0.
lb 4
6. x 4  3 x 2  75 x  10000 = 0.
7. a; 5 + 2 x 4 + 3 x 3 + 4 a; 2 + 5 x  54321 = 0.
NEWTON'S METHOD OF APPROXIMATION.
552. Find two numbers, one greater and the other less
than a root of the equation (Arts. 517 or 521). Let a he one
of those numbers, the nearest to the root, if it can be ascer
tained. Substitute a + y for x in the given equation ; then y
is small, and by omitting y 2 , y 3 , , a value of y is obtained,
which, added to a, gives b, a closer approximation to the value
of x. Now substitute b + z for x in the given equation, and a
second approximation may be obtained by the same process as
before. By proceeding in this way, the value of the root may
be obtained to any required degree of accuracy.
The assumed value of x should be nearer to one root than
to any other, in order to secure accuracy in the approxima
tion.
EXAMPLES.
1. Find the real root of the equation x 3 — 2 x — 5 = 0.
When 2 and 3 are substituted for x in the equation, the re
sults are — 1 and + 16 respectively ; hence a root lies between
2 and 3, and near to 2. Substitute 2 + y for x, and there
results
y 3 + 6y" + 10yl = 0.
Whence, approximately, y = .1.
Now substitute 2.1 + z for x, and there results
.061 + 11.23 2 + = 0.
420 ALGEBRA.
Whence, approximately, z = ' i}0 = — .0054, and
x = 2.1  0054 = 2.0946, nearly.
Find one root of each of the following equations :
2. a; 8 — 3 x ( 1 = 0. 3. x s  15 x + 03 x  50 = 0.
ANSWERS TO EXAMPLES.
In the following collection of the answers to the examples and problems given in the
preceding portion of the textbook, those answers are omitted which, if given, would
destroy the utility of the problem.
Art. 47; pages 10 and 11.
1. 93. 5. 408. 9. 5§. 13. 36. 17. llf.
2. 136. 6. 254. 10. 13$. 14. 48. 18. 9.
3. 127. 7. 24. 11. 4. 15. 3. 19. 10.
4. 156. 8. 310. 12. If. 16. 4. 20. 76.
Art. 60; page 18.
6. Ua9vip 2 . 7. x. 8. Sablcd. 10. 3mn 2 2x 2 y.
11. 39 a 2 2i ab + 5 h\ 12.  « + 3 c + 2.
13. xy+3m + 3n. 14. 3a + 3b + 3c + 3d. 15. x.
16. n+ r. 17. 6mn — ab — 4 c + 3x + 3m 2 — 4 p.
18. 4 a  2 b  12  3 c  d + 4 x 2  18 m. 19. 6 a 3 .
20. 14 six. 21. 7 a b + 7 (a + b). 22. 16 \/ y  4 (a b).
Art. 66; page 21.
6. — 3a5+4ctf— 5 a x. 7. 6 z + 12 ?/ — 8 « + 4.
8. _4a6c14a2y 148. 9. 2 y/ a  4 y + 12 a .+ 1.
11. 14 x 2 Sj/ 2 + 5ab7. 12. 2 b 2 e. 13. 6 b + 1.
14. 4m^8»r+3s. 15. Qd 2 b — 3 a 3c
16. 5m 2 +9w 3 71a;. 17. 2 b. 18. a 6 3 c.
422 ■ ALGEBRA.
Art. 74; page 24.
4. a — b + c + cl — e. 5. 2 a + 2. 6. x — y. 7. a — 3b + c.
8. 5wr6»4(i. 9. 6?» — 3%. 10. 4x + 2y.
11. —35 — 7c. 12. a — c. 13. 9a+l. 14. 6 m + 2.
Art. 86; pages 29 and 30.
3. 6a 3 16a 2 ?/+6ay 2 + 4?/ 3 . 4. a 4 + 4a + 3. 5. a 2 b 2
+ 2bcc\ 6. 0a 2 +16a&8£ 2 . 7. 6 8 a 8 . 8. o 4 «.
9. 30 a 3  43 a 2 b + 39 a b 2  20 b 3 . 10. 6 a 4 + 13 x 3  70 a 3
+ 71a20. 11. a 5  37a; 2 + 70 x 50. 12. Gar 5  25 a 4
+ 7a 3 + 81a 2 + 3a28. 13. 2a 5 b 2 3a 4 b 3 7 a 3 b 4 + 4:a 2 b s .
14. 4x 2m+1 y 3 — 1G x m+6 y n + 1 + 12 x 5 y 2n ~\ 15. 12 a; 6 +7 a; 4
+ 5 x 3 + 10 x — 4. 16. m 5 + re 5 . 17. a 5 — 5 a 4 6 + 10 a 3 b 2
10 a 2 b 3 + 5 ab 4  b\
Art 87; page 30.
2. 6 a 2 + 11 a J + 4 6 2 . 3. a 5 + x 5 . 4. a 8  2 a 4 a 4 + x 8 .
5. 2 a m + 1  2 a n + 1  a m + n + a 2n . 6. 1  a 8 . 7. a 3 + 3 a 2 a
 10 a x 2  24 x 3 . 8. a 5 5 a 4 + 10 a 3 10 a 2 + 5 a 1.
Art. 101; pages 37 and 38.
3. a x  2. 4. 3 b 2  4 a 2 . 5. 4 « 2  3 b 2 . 6. 3« 4 + 3a 3 i
+ 3a 2 6 2 + 3«6 3 + 36 4 . 7. a 2 ax + x 2 +^—. 8. x 3 x 2 y
a + x
+ xy 2 y 3 + —¥—. 9. 2 x 2  7 a  8. 10. 5 x 2  4 a + 3.
a; + y
11. x 2  2 x  3. 12. a 4 + x 3 y + a 2 ?/ 2 + x y 3 + y\
13. 3a 3 2a 2 + a5. 14. 2 a 3  a + 1. 15. ab+c.
16. a 2 — 3 a— y. 17. x + //. 18. a n — b m + c r .
19. l + 2a + 2a 2 + 2a 3 + ... 20. aax + ax 2 ax 3 + ...
ANSWERS TO EXAMPLES. 423
21. a*an + a i b 2 ab* + b i . 22. 2 a 3  2 a 2  3 a  2.
23.  x 2  2 x  4. 24. a 3  a: + 2. 25. 2 a 2  a b + 2 & 2 .
Art. 107; page 40.
23. 1  a 2 + 2 a b  b 2 . 24. a 2  b 2  2 b c  c 2 .
25. a 2 2ab + b 2 c 2 . 26. c 2  a 2 + 2 «6 £ 2 .
27. a 2 + 2«i + i' 2 c 2 + 2^ c/ 2 . 28. a 2  2 a 5 + 5 2
c 2 +2crf£ 29. a 2 + 2 a & + & 2  c 2  2 c d  d 2 .
Art. 115; page 42.
3. ( a + x )(b + y). 5. (x + 2)(x y). 7. (x 2  y 2 ) (m  n).
4. (am)(c + tf). 6. (ab)(a 2 + b' 2 ). 8. (a: + 1) (x 2 + 1).
9. (3 a; + 2) (2 a; 2 3). 12. (abcd) {ac + b d).
10. (2 a;  3 y) (4 c + d). , 13. (m 2 xny) (n 2 x  m y).
11. (27m 2 j(3»4m). 14. (4ww7a;y)(3a& + 5crf)
Art. 117; page 45.
9. (a + b + c + d) (a + b — c — d).
10. (a — c + b) (a — c — b). 11. (m + x—y) (m — x + y).
12. (x — m + y — n) (x — m — y+ n).
16. (x + y + 2)(x + y2). 18. (3c + d+l)(3c + dl).
17. (a + bc)(ab + c). 19. (3 + x 2 2y) (3x 2 + 2y).
20. (2 a — 5 + 3 d) (2 a  J  3 d).
21. (2 m 2 + 2 b  1) (2 m 2  2 b + 1).
22. (a — m + b f n) (a — vi — b — n).
23. (a + m + b — n) (a + m — b + n).
24. (x — c + y — d) (x — c — y + d).
Art 118; page 49.
25. (x 2  24) (x 2  5). 27. (x y 3 + 12) (x y 3  10).
26. (c 3 + 11) (c 3 + 1). 28. (a b 2  16) (a b 2 + 9).
424 • ALGEBRA.
29. (x + 20 n) (x + 5 n). 32. (x + y  5) (x + y — 2).
30. (m 2 + 11 n 1 ) O' 2  6 ir). 33. (x  8 y 2 g) (aj + 6 y 2 s).
31. ( a _ 5 _ 4) ( a _ 5 + 1). 34. (m + n + 2) (m + ra  1).
Art. 121 ; page 53.
3. Sab (a ' r 2) 2 . 7. 3 a 2 (a  5) (a  2).
4. 5 x y 2 (3 x  4 y 2 ) 2 . 8. 2 c m (c + 7) (c  3).
5. 2ay(3a; + y)(3a; — y). 9. a; y (m — 6) (m + 2).
6. a; (x + 7) (a; + 1). 10. ±ab(2a + b) (±a 2 2ab + b 2 )
11. (n  1) (>r + » + 1) O 6 + w 3 + 1).
12. (x 2 + y 2 )(xt 2 /)(^ 2 /).
13. (cc 4 + m 4 ) (a; 2 + m 2 ) (x + m) (x — m).
14. (m + w) (??i — w) (m 2 + m w + ?z 2 ) (??i 2 — mn + ri 2 ).
15. (a + c) (a 2  a c + e 2 ) (a 6  a 3 c 3 + c 6 ).
16. (2 a + 1) (2 «  1) (4 a 2 + 2 a + 1) (4 a 2  2 a + 1).
Art. 125 ; pages 55 and 56.
Z. ax. 6. x + 7. 9. x (x — 1). 12. 2 a: + 5.
4. m + n. 7. 2 a; — 3. 10. a — 2 &. 13. m(j; 1).
5. x 2 + 1. 8. 3 a  4. 11. x + 6. 14. 4 a?  1.
Art. 126 ; page 61.
6. 2 x + 3. 10. 2 a;  5. 14. x 2 + x + l.
7. 8a; — 7. 11. 5x + 3. 15. a — a;.
8. x  1. 12. jb + 2. 16. x 2  2.
9. 3 a; + 4. 13. 2 a;  1. 17. 2 (x + y).
18. 2«3x. 19. 3 a; + 2.
Art. 130; page 63.
2. 120a 4 6 2 c. 4. 36 a & b\ 6. 840 a 2 c 2 d 3 .
3. 30xV« 8 . 5. 480 m 3 rt 2 x 2 t/ 2 . 7. 252 a 8 y 3 s 8 .
8. 1080 a 2 b 2 c 3 d\ 9. 168 m n 2 x z y s .
ANSWERS TO EXAMPLES. 425
Art. 131; pages 63 and 64.
2. ax(x\a)(x—a)(x+ax+a 2 ). 7. ax (x — 3) (x— 7) (x + 8).
3. 12 abc (a + b)(ab). 8. (2x+l)(2xl)%4:x 2 +2x+l).
4. «(cc + l)(a;l)(x 2 — rc + 1). 9. 3 ab (x — y) 2 (a — b).
5. 24(l + cc)(lx)(l + a: 2 ). 10. 2az; 2 (3;c + 2) 2 (9a; 2 6x+4).
6. (x+l)(a;2)(cc+3)(*+4). 11. (xl)(x3) (x + 4)(x5).
12. (x + y + z) (x + y — z) (x — y + z).
Art. 132; page 65.
2. (3x4)(4 : r5)(2a; + 7). 4. (a 2 2a2)(a + 3)(2al).
3. (4a;+l)(2a;+7)(3a;8). 5. (2x + 3)(x 2 x + l){x 2 +x2).
6. (a — b)(a 2 ab + b 2 )(a 2 + 2ab + b 2 ).
7. a x (x + 1) (x 2 — x — 1) (x 2 + x + 1).
8. x (x  5) (2 x 2  x 2) (3 x 2 + x  1).
If the above expressions are expanded, the answers take the
following forms :
2. 24 z 3 + 22 a 2 177 a; + 140. 4. 2« 4 + a 3 17« 2 4« + 6.
3. 24 z 3 + 26 z 2  219 a; 56. 5. 2x 5 + 3x i ±x 3 + 5xQ.
6. a 5 a 3 b 2 + a 2 b 3 b\
7. a x G + ax 5 — ax 4 — 3 ax 3 — 3 ax 2 — ax.
8. 6ic 6 31a; c 4x 4 + 44a; 3 + 7a; 2 10a;.
Art. 148 ; page 71.
14  *£■■ 18 
o + 2 c
a (2 + 3 n)
1D * b{23n) • 19 '
16. ±*J*y+v\ 20 .
m + 9 3 y — 5
10.
cd
3 xy'
11.
x a
2?'
12.
sc — 5
# + 7'
13.
m — 2
2
x 2 y
5
— X
2 + x
X
c
(Jx)
d
c
III
+ d'
. — nr
m 2 — n
426 ALGEBEA.
Art. 149; page 72.
2. 3 *
3.
4 a; + 1
5a + 7
a2 '
10.
4.
m — 1
6 m — 5
5.
x + 2
x — 3 "
x'
2
3a; + l
x' z — x + 3
Art. 150; page 73.
6.
3 a; — 2
8 2a; + 5
2 x 7
a; + 3 *
7.
2x3
_ C « — 1
"• 7= »
5a — 7
2a;5*
, 1 2x 2 a;2
U< 2 a 2
+ 3a; + l*
3
a 
a*
4.
a; 2
— X
!/ + */•
6.
a; 2
3"
a;
~~3
7
+ 3"
2
a;"
7.
a
2b'
3 2b
2 + a '
K 2x 3
o
5 5 5a;
23
8. 2a; + 6 +
9. x 2 + a; + l. 10. 2 + 77 ^ — ^, 11. a;24
12. jc
x — 3
2a;4
2x z — x+1 ,x z +x — 1
a; — 2
2a; 2 3a; + 3'
Art. 151; page 74.
_ (x — 1) 2 _ an + b—cd 56.r — 4?i 2 — 5a
2. =— . o. . 4.  .
x — 6 n o
(a; + l) 2 . 2ab „ a 2 +2b 2 Q a 3 + b 3
0. . b. j. 7. — s . o. — .
x a+b 2 a a — b
Q 6a; 2 7a;l 2b 2 ,, a; 3 2a; 2 3a;
"• ^ : n • I". — ■ . 11. „ .
2a; — 1 a+& x — 2
Art. 152; pages 76 and 77.
27 a b 16 a c 30 6 n 3 a; 2 ?/ 2 a?/,? 7 y z 2
3 ' 772~' "T^ 7 ~72~" 30 ' 30 ' 30 '
18 ?/a 2 16 x 2 z 2 15 a; 2 y 2 '
12 xyz* 12xij z' 12 xyz'
40c 2 10c 18i 2 12 6 25 « a _2^_ 3_o^ 4 a a; 2
30 a&c ' 30 abc ' 30 a be' ' a*x 8 ' a 8 a; 8 ' a 8 a; 3 '
8.
ANSWERS TO EXAMPLES. 427
100 a y z 3 Aob x 3 z 8Acxy s — 12mx y 2
120 x 1 ?fz 2 ' 120 x 2 y 2 z 2> ' 120 x 2 y 2 z 2 *
(a+b) {a 2 + U 1 ) (a  b) (a 2 + b 2 ) a 2  b 2
10.
a 4 6 4 atb* ' atW
x 2 — 9 x 2 l x 2 4
(xl)(x2)(x3)' (xl)(x2)(x3)' (xl)(x2)(x3)'
2 a (a + 2) 3 b (a — 2) 4 c (a + 3)
12.
02)(a+2)0+3)' 02)0+2)0+3)' (a2)(a+2)(a+3)'
x 3 +2x 2 \2x + l x 2 + x + l x + 1
(x + 1) {x 3  1) '0 + 1) (X s  1)' {x + 1) (a 3  1)'
13 6 a 2 b 2 3 b (m 2  n 2 ) 2 a (a 2  b 2 )
6ab(a — b)(m + ?iy 6ab(a—b)(m + n)' 6ab(a—b)(m + ii)'
3 Q+l) 2Q1) 2a 1x x 2 x2 3
' a»l' a 2 l'a 2 l' "*' lx» 1jb 2 >1=?'
17 c2 ^ 2 Ql)Q + &) Q6) 2
' O 2  6 2 ) (c  d)' O 2 * 2 ) (ed)' (a 2 b 2 )  d)'
Art. 153; page 78.
2 ( a ~ b y 3 g 2 +9«+8 9 m 2  4
a* — b 2 ' ' x 2 +5x — 24' "6m 2 — 19m + 10"
4 2 +«& + 6 2 ) 1_^
a 3 — & 3 1 — x
Art. 154; pages 80 to 82.
4 12 a; + 7 6a + 5& a + 3 3 m 2 n 2  4
36 ' ' 10 a 2 6 2 ' 24 • 6m 2 » 3 *
8 5 &* + 4 «' q 5a + 6 in 4,abb±a 3 1
" 120 a 6 " * 24 • iUl 12« 3 T~ ~* 1L 15"
12 — is 3a; ~2 1 45ctf+6acd— 3ahd— 2abc
•42' "■ 18 x 2 ' 14l_ 60 15 ~ ~48^X~
17. ? 18 1 19 2 Q^ + & 2 )
6 + xx 2 ' ' x 2 +15x+56' a 2 b 2 '
428 ALGEBKA.
■Oi *" «l£±. 22.. 4^. 23. (* + 2 y
1_^' — a6* a 4 ^' • (.x + l)(a; 3 l)
1318* _J_ a 2 14a + l
^ (z + lXa^XzS)* 0, &«' ' 6(^1)
28. 57 _?_. 29. 44 30. 0. 31.
9 a: a; 3 ' ' »*— 1" ' ' ' (»2)(jb3)(»4)
Art. 155; pages 83 to 85.
2.
a 5 b s c
m* n 3 d'
3.
12 a 4 b x
35 h 5 m '
4 ^
7.
/ • 8 
4 a.' y
a 3 .
9 11
o
m
4
w 2 3 a;
a; 
1 10*
2 * "■ 3
13.
b(ab)
x(a + b)
14.
a — b
a 2 '
I*' 1 '.
a;
ia x*x2Q
16 '
17.
a (x2)
a + 1
18.
X
x2'
19. "+*•
. or
20. aj2 + 5 f + 6
ST
21. ^o . 22. x + xy. 23. 1. 24. 2. 25. J£j.
ic + 2 % ~r y
. 91 ra 2
o.
6rc 2
2
4.
Art. 156
; page 86.
5 mx
7.!.
X
9.» + 1 '
a + 5
, 3 04)
8 3.t 2?/
10. aj.
8 b 3 m ri 1 x s x + y
Art. 157; pages 88 and 89.
a
_ acn + bn „ 3m— n „ Ay—±x + 2a
— — . o. — — . b. — x . «. ^j •
bm + b?i cnx — cm ox oi
x 2 y 3 +l
8. x — 1. 9. a; 2 rK + l. 10. a + b. 11.
xif — ly*
, n a— J 10 «— 4 ... 1( . 4 R aft
12. T7. 13. ^. 14. aj. 15. k— Q lb  ■> , 7,2 ■
a b x + 6 3 a; + 3 a" +
ANSWERS TO EXAMPLES. 429
17 1 18 ww(mw) 8 19 _ sg
m* + m 1 ri 1 + w 4 ' x + 2 a
Art. 175; pages 94 and 95.
2. aenx—bcen=bdnx—bem. 3. 6 bx — 8a 2 = 3 — 2a6a.
4. bdex— adex+bcex— abd=0. 5. \2x\hx — 6x— 1320
6. 9z12a=10a:+2446. 7. 28a:4a;+560=14a;+7a;+728.
8. ±ax6c5a s x + 2a 3 bd=0. 9. 10x32a;312=2152a;.
11. 3 a — 2 a — 2 a; = 45. 12. a 6 a; + b 1 — ex — d = a c.
13, 33z22x = 0. 14. 6ar+3z6ar+ 184a;2=0.
15. 3 a ._3_2a25x=0. 16. 6x + 615aj + 4520a;10=0.
Art.
177; pages
97 to 102.
4.
3.
13. 1.
22. 72.
31. 5.
41.  1&.
5.
7.
14. 2.
23. 60.
32. 5.
42. 4£.
6.
1.
15. 2.
24. 10.
33. 4.
43. l t V
7.
16. 4.
25. 2*
34. 5.
44. 0.
8.
1.
17. 2.
26. 56.
35. 2.
45. H.
9.
3
<
18. 1. >
27. ?.
33. f .
46.^.
11.
3
— 2'
20. 3.
29. 2.
37. 1.
47. 1.
12.
0.
21. 5.
ao.J.
40. 7.
50. f *
2a+b
51.
a 2
a 2 
+ 4a . 52
3a+2
!. 2&. 53.
If *I
Abc + a*
a?—b + 16c
55.
2 a 2
36
, 57. ^.
59. \
a + 2
i 61 ^
64. 3.
56.
a.
r
58. 12 a 3 .
60. aft.
63. 2.
65. 50.
66 B5
67. 5.
68. 0.
430 ALGEBEA.
Art. 182 ; pages 108 to 113.
10. Horse, $224; chaise, $112. 11. 37. 12. 10 and 7.
13. 18 and 2. 14. 58J and 4l£. 15. A, 40 ; B, 20.
16. A, 60 ; B, 15. 17. If. 18. ft. 19. 23£.
20. 84. 21. 36. 22. Oxen, 12 ; cows, 24.
23. Wife, $864; daughter, $288; son, $144.
24. Worked, 20 ; absent, 16. 25. Horse, $ 126 ; saddle, $ 12.
26. Infantry, 2450 ; cavalry, 196 ; artillery, 98.
27. 144 sq. yds. 28. Water, 1540 ; foot, 880 ; horse, 616.
29. $ 1728. 30. $ 2000 at 6 p.c. ; $ 1200 at 5 p.c. 31. 7.
32. 31. 33. $24. 34. $100. 35. 142857.
36. A, $ 466 ; B, $ 533J. 37. 2 dollars, 20 dimes, 4 cents.
38. $2.75. 39. Men, $25; women, $21. 40. 23 and 18.
41. 48 minutes. 42. 12121 men ; 110 on a side at first.
43. 5 r \ minutes after 7. 44. 43 T 7 T minutes after 2.
45. 27f\ minutes after 5. 46. 29 and 14.
47. 3377 ounces of gold ; 783 ounces of silver. 48. $ 2000.
49. 30 bushels at 9 shillings ; 10 at 13 shillings. 50. 10 a.m.
51. $ 1280. 52. 21 ft minutes, or 54 T <Y minutes after 7.
53. 27^ T minutes after 4. 54. 23*1 miles.
55. Greyhound, 72 ; fox, 108. 56. 1 minute, If §f seconds.
Art. 192 ; pages 120 to 123.
3. x = 4, y = 3. 9. x=2, y=10. 15. x = 12, y = 18.
4. s = 5, y = 2. 10. jk==12, y = S. 16. a; = 35, y = 10.
5. x = 7, y = 5. 11. x=~2, y=10. 17. x =  28, y = 21.
6. x =  8, y = 2. 12. x = 10, y = 5. 18. x = A,y = .1.
7. x = 5, y = 7. 13. x = 7, y = 11. 19. x = l£, y = 3f .
8. x=8, y=12. 14. a>=ll, y=9. 20. x = 3, y = 2.
21 dm — bn _an — c m n / + ri r
ad — be ad — be ~ mri + m' n '
ANSWERS TO EXAMPLES. 431
m'r—mr' nn ac(bvi + dv) bd(cnam)
V = • "3. X = ^= ; , y = ^ : .
u m n' + w! n ad + b c ad + b c
11 25 5*
24. x=—, y=^. 25. x = 60, y = 40. 26. x = — , y=».
2 a 2, a Ob
27. x = lft, y = 4 T V 28. x = 6, y=5. 30. # = 4, y=2,
be — ad
bn — d vi '
31. x = — 5, y=3. 32. x = 2, y — — l. 33. ■ x =
b c — a d „. 3 2 oe 1 1
y = • 34.33 = ^77,2/ = —^. 35. x=,y = —
cm — an a~ b a tr n vi
Art. 194; pages 126 and 127.
3. x = 23, y = 6, g = 24. 6. a; = — 5, y = — 5, z = — 5.
4. x. = — 2,y = 3, z = l. 7. « = 4, x = 5, y = 6, z = 7.
5. x = 8, y =  3, g =  4. 8. a; = 3, y =  1, s = 0.
9. jc =^ (b + g — a), y = ^ (a + c — b), y = ^ (a + & — c).
10. x~^,y = 7 £,z=^ 11. a = 24,y=48,g=60.
12. u = — 7, x = 3, ?/ = — 5, z = 1.
_ 5 2 +c 2 ft 2 _a 2 +c 2 & 2 _ a 2 + & 2 c 2
13, *~ 2bc ~ ,V ~ 2ae ,Z ~~ 2ab '
14. x = ,y = ^,z=. 15. a; = li y = X\, «='l.
16. x = ab c, y=ab + ac + bc, z = a + b + c.
a + 1 f
17. x = 7, y — — 3, g = — 5. 18. as = ,y—a — c,z
c a
Art. 195 ; pages 129 to 133.
4. A, 30 ; B, 20. 5. ^. 6. Cows, 49 ; oxen, 40.
7. A, $140; B, $70. 8. A, 98; B, 15. 9. 32 and 18.
10. Man, 24 ; wife, 18. 11. Worked, 6 ; absent, 4.
12. Horse, $96; chaise, $112. 13. A, $96; B, $48.
432 ALGEBRA.
14. 16 days. 15. 13J bushels at 60 cts. ; 26 at 90 cts.
16. Wheat, 9; rye, 15. 17. Income tax, $20; assessed tax, $30.
18. A, $500; B, $700. 19. 30 cents ; 15 oranges.
20. 1st, 8 cts. ; 2d, 7 cts. ; 3d, 4 cts. 21. Better horse, $40 ;
poorer, $30; harness, $50. 22. 10, 22, and 26. 23. 246.
24. A, $2000; B, $3000; C, $4000; D, $5000.
25. A, 45; B, 55. 26. A, $20; B, $30; C, $40.
27. Whole sum, $120; eldest, $40; 2d, $30; 3d, $24; 4th, $26.
28. Length, 30 rods ; width, 20 rods ; area, 600 sq. rods.
29. Going, 4 hours ; returning, 6 hours.
30. A, 9f days ; B, 16 ; C, 48. 31. 1st rate, 6 p.c. ; 2d, 5 p.c.
32. 15 miles ; 5^ miles an hour. 33. 30 miles an hour.
34. A, 5 ; B, 6. 35. First, 22 ; second, 10. 36. A, 8 ; B, 6.
Art. 197; pages 136 and 137.
. a b c _ . , _ m a .. n a
4. — — . 5. li hours. 6. and .
ab + ac + bc vi\n m,\n
7. 12 and 8. 8. ^. 9. 12. 10. 10 °" 11. $2100.
b — a rt+ 100
12. 100 >i>), 13. m. 14. lst, g(c  6) ;2d ; ft(a  r) .
p r ~ a — b a — b
,..». ^,.,/n , „ b + d ^„ am+bji + cp
15. 1st kind, 5 ; 2d, 10. 16. — — . 17.  r — .
a — c a + b+c
* amt . o ant ' n a P t "
mt + nt' \pt' n ' mt + nt'+ptf n ' mt+nt'+pf'
Art. 205; page 141.
5
3.2 rods. 4.  ^ . 5. 105 and — 15. 6. In — 30 years.
7. A, .$1500; B, $500; that is, A was in debt $1500,
and B $ 500. 8. Man, $ 3 ; son, — $ 0.50 ; that is, the man
was at an expense of 50 cents a day for his son's subsistence.
ANSWERS TO EXAMPLES. 433
Art. 225; page 152.
4. x > 5. 5. x > 15, x < 20. 6. 4. 7. x > 6, y > 2.
8. a; > c, x < d. 9. * > 9, 7/ < 12£. 10. 19 or 20.
11. Any no,, integral or fractional, between 8 and 15. 12. 60.
Art. 229; page 155.
2 7 2
1. a 3 3a 2 b + 3ab 2 b 3 . 2.~2+— i .
3. 1 + 3 a 2 + 3 b 2 + 3 a 4 + 6 a 2 6 2 + 3 b* + a & + 3 a 4 b 2 + 3 a 2 6 4 + 6 6 .
4. a 2 + 2am — 2an + m 2 — 2mn + n 2 .
5. a *m _ 4 a 3m + », __ (5 a 2m + 2n _ 4 (( m + 3a + a 4 n#
6. a 5 + 5 a 4 5 + 10 a 3 b 2 + 10 a 2 b s + 5 a 6 4 + 6 5 .
Art. 230 ; page 156.
3. 4a: 4 +12ar+25ar 2 +24z + 16. 4. 4a: 4 12a; 3 + lla; 2 3a: + :.
6. a; 6 +4a: 5 +6a; 4 +8a; 3 +9a: 2 +4a;+4. 7. l4a;+10a: 2 12a; 3 +9a: 4 .
8. 1 + 2 a; + 3 ar + 4 x 3 + 3 a; 4 + 2 a: 5 + a: 6 .
9. x 6 8x 5 + 12 x* + 10 a; 3 + 28 x 2 + 12 x + 9.
10. 4 x« + 4 a: 5 + 29 x" + 10 a; 3 + 47 a: 2  14 x + 1.
11. a; 6 + 10 x 5 + 23 x i  6 a: 3 + 21 x 2  4 a; + 4.
12. 9 a; 6  12 x 5  2 x i + 28 x 3  15 a; 2  8 a: + 16.
Art. 231; page 157.
2. a^+Qa'b + 12 a 2 b 2 +U 3 . 3. 8m 3 +60?7r?z+150mw 2 +125» 3 .
4. 27 a; 3 108 a; 2 +144 a; 64. 5. 8 a; 9  36 a; 6 + 54 x 3  27.
6. 64 x 6 48 x 5 y +12 a; 4 ?/ 2 — a; 3 ?/ 3 .
7. 27 a; 3 ?/ 3 + 135 a b 2 x 2 y 2 + 225 a 2 b A xy + 125 a 3 b 6 .
Art. 232; page 158.
3. a; 6  3 x 5 + 5 a; 3  3 x  1. 5. 8  24 a; + 36 x 2  32 a: 3
+ 18 x i  6 a; 5 + a; 6 . 6. 1 + 3 x + 6 a; 2 + 10 a; 3 + 12 a; 4
434
ALGEBRA.
+ 12 x 5 + 10 x« + 6 x 1 + 3 x 8 + x\ 7. 8 x»  12 x s + 30 a; 7
 61 x* + GG x 5  93 a 4 + 98 a; 3  63 a 2 + 54 a;  27.
Art. 239 ; pages 162 and 163.
2. 2 a; 2  x  1. 5. 3  2 a; + a; 2 . 8. 3 ar  4 x  5.
3. 2 a 2  4 a + 2. 6. 5 + 3 x + x\ 9. 2 x 2  5 a; + 8.
1
4. m + 1
m
7. 1 — 7» 2 a; 2 . 10. a b
c.
U. x2y + 3z.
13. a+ S fr—, : 4
a;
a; 2 a; 3
i 3
8 a 3 16 a 5
15. a + s
12  1 + 2^ + 16
2 a 8 a 3
+
x°
16 a 5
Art.
241; page 166.
2.
523.
7.
95'
12. 900.8.
17.
13.15295
3.
214.
8.
1.082.
13. .4125.
18.
.88192.
4.
327.
9.
21.12.
14. 1.41421.
19.
.43301.
5.
5.76.
10.
.083.
15. 2.23607.
20.
.57735.
6
.97.
11.
.00328.
16. 5.56776.
21.
.53452.
1. 3.3166.
2. 1.732051.
Art. 242; page 168.
3. 7.81024968. 5. 27.94638.
4. 11.446. 6. 113.7234.
Art. 243; pages 170 and 171.
2.12?/. 4. 4:x3ab. 6. y 2 — y1. 8. a; 2 2a; + l.
3. 2ar + 3. 5. a: 2 + 2a;4. 7. x + .
x
9. a + b + c.
10. 2 a; 2
3a;l.
115
11. X+'„ oK— R +
3 a; 2 9 a; 6 ' 81a; 8
12.
x
ANSWERS TO EXAMPLES.
435
a?
a G 5 a? g 2 1 1
5
3x 2
~9x 5 Six 8 '" . 4z 4 32z 10
768x 16 "*
Art. 245; page 173.
2. 123. 5. «. 8. 1.442. 11. .855.
3. .898. 6. 3.72. 9. 1.913. 12. .420.*
4. 11.4. 7. .0803. 10. 5.963. 13. .561.
Art. 247; pages 175 and 176.
2. to 2 — 2m — 4. 3. a 2 ax + x 2 . 4. 2x — 1. 5. x' 2 — x + 1.
Art. 248 ; page 176.
1. 2x3y. 2. cr1. 3. to 2 2to3.
Art. 257 ; pages 180 and 181.
4. c^. 5. af*. 6. *»*. 9.  6 a J*K
11. a 4 6~ 4  2 + a 4 i 4 . 12. a  b.
13. a 5 3a~ 3 b 2 + a 2 b s 2a 1 b i . 14. 18a 2 i 2 +10 + 2« 2 & 2 .
15. 2 a r 1 2/10z Z / 1 + 8.ry 3  16. 24x"V + 2affy.
17. 6x 2 7^19af*+5a;+9^2^. 18. 32aZr 2 50+18 a  1 & 2
Art. 258; pages 182 and 183.
5. <T*. 6. to 5 . 7. x 12 \ 8. ^ '
11. J + Jb^ + Jb^ + Jb^ + bl 12. a^^^ + i 2
13. x 3 f x~ 2 y + 2x\ 14. a>* y 1  3 + 4 x~* y,
15. x 1 y 2 a; 2 7/ 3 x 3 y 4 . 16. 2x^y~ ?J x~^yx~^ >*.
Art. 260 ; page 184.
6. x*. 7. c~K 8. to 1 . 9. 2T 3 . 10. a~ 3 . 11. w" 1
436 ALGEBRA.
Art. 262; page 186.
' 1 n 12500
3  9  5 ioooo 7  4 ' 9± ^~
4. ±216. 6. ±^. 8.243. 10. ±£.
Art. 263 ; pages 186 and 187.
5. 3 32 y _ 2 x 1  2T 1  6. 2 aj^ + x y~^ 4^ y~*.
./•'
7. a?t"y T *2+aT*y*. 1L 2y*y*ar 1 . 12.
13. a; 2 " 6 . 14. a*». 15. a 31 . 16. a  ^. 17. sc.
01 •* + «* oo h^d'afd* 9q 5 »" (»»!)
21 1_3« 3 * ^ ab*# + aP*' 3a '
Art. 267; page 189.
2. ^27, #16, #25. 3. $625, ^216, ^49.
12.^ 12, 12 „ 15,— — 15,——— 15,.
4. V*V, V^ 4 ^ VVs 3  5. #32 a 5 , #27 & 3 , #64c 3 .
12. 12,
6. #cr+2a6+6 2 , \/a 3 3ab+3ab 2 b 3 . 7. #a°3a 4 .z 2 +3a 2 ;c 4 x 6 ,
fa 8 2« 3 x 3 + x 6 . 8. y^3. 9. #2. 10. #4.
Art. 269 ; page 190.
5. \l$ab\ 6. #a& 2 . 7. v/(f 3 )
Art. 270 ; page 191.
11. 3xyS/2xy' 2 3xhj. 12. (a — 3)^a. 13. (x + y)^x — y.
14. (2 x + 3 a) f 6a. 15. 4 a & #3a& 2 + 5&. 18. ^ 6 
1 ._ „„ 1 ,„. „. 2a
19. ^i/30. 20. Ji/21. 21. ^f y/3. 22. r> #6*.
6 b y J
ANSWERS TO EXAMPLES. 437
a\j ab c
2 (a + x) 6 (a + 6)
Art. 272; page 192.
7. v^x s ft^F 9 y/(^)
Art. 273; pages 193 and 194.
«/. o 38
3. 10 si 2. 4. 12 v/ 3. 5. 9^2. 6. ^y/o
7 \/6. 8. j2 + ^18. 9.4^5. 10. i^
11.6asJ3a. 12. ^V 3  13. ^2 + ^3.
14. 2 yV 2  tf. 15. (2 a  5 b) \Jl x.
Art. 274; pages 195 and 196.
5. \f^¥x\ 6.^4500^ 7  ^(g^ga)
8. y/5^ . 10. a; + v/z  6. 11. 21 x  38 v^ + 5.
12.2. 13.1. U. xijz + 2\jy^.
15.4 + 2^10. 16. 56+12v/35. 17.3632^15.
18. axx\ 19. m + TO. 20. 14 — 4^6.
21. 147 + 30 v/24. 22. 1 + 2 a %/la 2 . 23. 2 a 2 \/a 2 & 2 .
Art. 275 ; page 196.
b/16
V / 243'
e/8
V9
7. {/" 8. {18. 9. ^
438 ALGEBRA.
Art. 276; page 197.
3. ^125. 4. y/7. 5. 2304 x\ 6. a 4 * 2 . 7. \j~a~^~b.
8. SlaHx Sjbx. 9. a: 2 + 2x + l. 10. 16 a: 2  48.
Art. 277; pages 198 and 199.
12,
3.^2. 4. si 2. 5.<)a + b. 6. V*  1. 7. y/2
8. ^3. 9. v/3. 10. ^^V n  V 2 
Art. 278 ; page 200.
. 3s/2 . ^4V 2 x 5 \/ 2 a 2°\/3tf
O. pr — . 4. — s . 0. — ^ — . D. 5
2 2a 2 3a
Art. 279; page 201.
3. 12 ~ 4y/2 . 4. 5 + 2 V 3. 5. 2 ^ 6  5.
a + 2_^b'+b 16 + 7 y/ 10
6 ' " «fi ^ ~* 7 * " 13
_ a — 2 ^ ax + x a + 3 + 3 \la + 1
a — x a
a+^a*x 2
10. 2a 2 l2aV / « 2 l H. • 12. y^lff 2 .
a;
1Q x*2 + x^x 3 4: 14 a:  24 11 Vs 8  2a;
2 18 — o x
Art. 281; page 202.
2. .894. 3. 7.243. 4. 3.365. 5. .101.
Art. 286; page 204.
4. — 8 v/6. 5. 12\/ab. 6. 46. 7. 2.
8. abc V^l. 9. a 2 + b. 10. 12. 12. sJB.
ANSWERS TO EXAMPLES. 439
13. y/2. . 14. y/5. 15. ^3. 16. \f^l. 17. 1 + <f^2.
18. 2 (f b ) . 19. 1 _ 4 y/^3. 20.  100  18 tf=2.
or + b T
Art. 293; pages 207 and 208.
5. v/T + y/5. 8. 5 + y/lO. 11.^15^5.14.32^2.
6. v/21v/3. 9. 3v/3. 12.3 + ^5. 15. 5 + 3 \/^2.
7. 3 + v/7. 10. v^5v/3. 13. 73 y/ 2. 16. 6Vl.
17. ^m + n— \jm — n. 18. x — SJax. 19. 3 + v/2.
20. v/21. 21. 2^3.
Art. 297; pages 209 and 210.
4. 17.
9. 4. 14. 4. 19. 1. 24. 4.
5. 19.
10. 5. 15. 81. 20. 3. 25. 5.
6. 7f.
11. 2. 16. 4. 21. 4. 26. 3.
7. 2.
12. . 17. 8. 22. 12. 27. 6.
8. 4.
13. 4. 18. 3. 23. 25. 28. 39.
29.
3£. 30. 3. 31. 6. 32. 3«l
Art. 303 ; pages 212 and 213.
2. ±3.
4. ±y/ (_?). 6. ±7. 8. ±1. 10. ±
3. ±5.
5. ±1. 7. iy/11. 9. ±1. 11. ±
Ifr — IA
12. ±^\L±y 13. ±v/a + ft.
Art. 310 ; pages 220 to 222.
10. 5 or  7. 11. 11 or  2. 12. 5 or 3.
440 ALGEBRA.
13.  5 or  13. 29.  4 or  1. 45. 2.
14. ^or. 30. 2 or i. 46. 4 or 0.
15. 2 or . ' 31. 4 or  If. 47. 3 or  2.
16,_l r. 32. 4±2y/3. 48. 2ov~
do bo
17. ^^^ 33.3orl. 49. ± 2
12 ■ — ~ .y/8'
18. 17± y 337 . 34. 2 or  1 . 50. 25 or 3.
4 7
19.  or 1 35. 7 or  . 51. 6 or 2.
o 2 6
20. lor 7 , 36. 4or^. 52. or.
4 4 a. c
21. =f or  2. 37.  10 ± v^78. 53. a ± b.
o
0Q 1±V409 3 a a
**. £ . 38. — 3i or — 2^. 54. — —  or  .
b ""42
23. — or  . 39. 1 or — . 55. — a or — b.
4 Z oh
24. 3£ or  1. 40. 1 or £ . 56. 11 or 18£.
25. 13 or  2. 41. 5 or ^ . 57. 5 or  3.
5
26. I or i . 42. 18 or 3. 58. 12± J .
2 14 5
27. 1 or 3i. 43. — 2 or — . 59. a — ft or — a — c.
28.  4 or  ** . 44.  3 or 2*. 60. ^=* or 3 ^.
_, a + b a — b
bl. or — — r .
a — o a +
ANSWERS TO EXAMPLES. 441
Art. 311 ; pages 224 to 227.
4. 12 rds. 5. 40000 sq. rds., and 14400 sq. rds. 6. 9 and 6.
7. 16 and 10. 8. 16. 9. 3 inches. 10. $ 30. 11. 14 and 5.
12. $2000. 13. 18bbls., at $4 each. 14. 256 sq. yds. 15.5.
16. 7 and 8. 17. 7, 8, and 9. 18. Length, 125 ; breadth, 50.
19. 9. 20. 3712. 21. 80. 22. 20.
23. Area of court, 529 square yards ; width of walk, 4 yards.
24. 36 bu. at $1.40. 25. Larger, $77.17^; smaller, $56.70.
26. 1st, 14400; 2d, 625 ; or, 1st, 8464 ; 2d, 6561. 27. 84.
28. 6. 29. Larger pipe, 5 hours ; smaller, 7 hours.
30. 38 or 266 miles. 31. 70 miles.
Art. 314 ; pages 230 to 232.
5. ±3or±V13. 6. ±l or ±J_ 7. lor 2.
I y 5
8. ± 1 or ±  . 9. ± 7 or ± 5. 10. ^3 or  ^23.
11. ± 8 or ± d( ^ . 12. 4 or ^49. 13. 4 or 1.
14. 243 or  J' (28 5 ). 15. 4 or 1\. 16. 49 or 25.
18. 2,  2, 3, or 7. 19. 3 or  1. 20. ± 1 or ± 2.
21. 2 or  3. 23. 1,  1, 5, or 7. 24. 2,  3, 4, or  5.
25. 1, 2,  5, or 8. 26. 1,  1,  6, or  8.
28. 3,or 3± 4 V/ 55 . 29. 8, 2, or 3 ± y/ 110.
80.?,?, or " 3± 2 2 ^ 3 . 31. 1,9, or 5±2 N /2.
32. 0,5, 1, or^.
Art. 317; page 234.
2. x = 2 } y—±l; or, x = — 2, y=±l. 3. x = 4, y = ± 5 ;
442 ALGEBRA.
„ . 1 1 11
or, x= — 4, y=±5. 4. « = g, y — ±^ ° r ' a;=— 3' ^ =± 2'
1 1
5. x = 3,y = ±p or,x = — 3,y = ±g.
Art. 318; page 235.
2. a? = 7, y = — 8; or, x = — 8, y = 7.
3. a; = 5, y = — 2 ; or, a; = — 2, y = 5.
4. x = 3, y = 4 ; or, cc = — 4, y = — 3.
5 5
5. x = S, y = 2> or > cc=: ~2' y:= — 8 *
1 5
6. x = 2, y = 4 ; or,x = — ^,y = ^.
7. ar = 2, y = — 3 ; or, x = 3, y = — 2.
8. x = 1, y = 2 ; or, a; = 2, y = 1.
Q Q 9 15 62
9. a? = 3, y — 2; or,x = — —,y = —.
10. a? = 9, y = 6 ; or, x = — 6, y = — 9.
11. a; = 2, y = 9 ; or, x = 9, y = 2.
12. a; = 9, y = 3 ; or, x = — 3, y = — 9.
13. x = 6, y = — 4 ; or, x = — 4, ?/ = 6.
14. a; = 3, y = 2; or, a; = —, y = — — .
15. x = 5, y = 3 ; or, a; = — 3, y = — 5.
16. x = 3, y = — 7 ; or, a; = — 7, y = 3.
Art. 319; page 238.
4. x = 3, y = 4; a; = 4, y = 3; x = — 3, y = — 4; or, x = — 4,
y = — 3.
U. & = 6, y = 7 ; a; = 7, y = 6 ; a; = — 6, y = — 7; or, a; = — 7,
y = — 6.
6. x = 2, y = — 3 ; or, x = — 3, y = 2.
ANSWERS TO EXAMPLES. 443
7. x = — 1, y = 4 ; or, as = — 4, y == 1.
8. # =1 3, y = — 2 ; or, x = — 2, y = 3.
9. a; = 4, y = — 7 ; or, a; = 7, y = — 4.
10. x = 5, y = 6 ; or, cc — 6, y = 5.
11. x = 5, y = 2 ; or, a; = — 2, y = — 5.
Art. 320 ; pages 239 and 240.
2. x = 2, y = 2> x =  2 ,!/ = 2> x = \ 5>y = — 2 \
2
5 ;
5 g
3. x = 2,y = 3; x = — 2,y = — S x = ^^ r ,y =
^31 ' y ~ v' 31 '
5 6
y/31' y "~^31.'
4. aj = 3, y = l; a>= — 3, y = — 1; x = 2^2, y = \/2;
or, ^ = 2^2, y = — ^2.
5. £ = 3, y = 5;a; = — 3, y = — 5; x = , y = y5 or, z = — ,
13
6. a; = 2, y = — l\ x = — 2,y = l, x = —r , y
5 7
7. a; = 2, y — 1 ; a; = — 2, ?/ = — 1 ; x = 7, y = — 19 ; or,
x — — 7, y = 19.
Art 321; pages 243 and 244.
5. x = l, y = 8; or, x = 8, y = 1.
6. a; = 4, y = 9 ; or, x = 9, y = 4.
7. a: = 2, ?/ = 3 ; or, a: = 3, y = 2.
444 ALGEBRA.
8. x=3, y=4 ; 3=4, y=3 ; x=± + \J^11, y =A\^H ■
or, x = — 4 — \/— 11, y = — 4 + y^— 11.
9. a:=4,y=5;a:=16, y=7 ; 3= 12+^/58, y=l— ^58 ;
or, x = 12  y/58, y =  1 + y/58.
10. a; = 4, y = 2 ; cc — — 2, y = — 4 ; or, a; = 0, y = 0.
11 o a 605 20 io , 3
11. x = 9, y = 4; or, *=.— , y = — . 12. a = 1, y = .
13. x = 3, y = 2 ; or, a; = 2, y = 3. 14. x — 9, y = 4.
15. 3 = 1, y=3; a=3, y=l; a=l+V^^2, y=l V^2;
or, a; = 1 — \/2, y = 1 + ^~2.
1« 1 o o , 3+V55 3+V55.
16. x=l,y=—2;x=2,y=—l;x= 1 ,y= j ;
or,x = , y = y. .
17 9 o o o 1+3V^3 1+3^.
17. x=2,y = 3;a:=3,y=2;a: = ^ ,y= J ,
13 y/^3 13 V / ' =r 3
or, a;= ^ » ? = ^ ■
18 . , = 3 ;y =2;, = 2 ,,=3 i ,=^±^,,==^. 9 ;
_9v/309 9 + ^309
12 " ,y ~ 12
19. 3=1, y=3; 3=l, y=3; 3=141, y=3f; or, 3=14f,
20. 3 = 2,y = 3; or,3 = 2f>y = lf.
01,100 4 22 59
21. a: = 4, y = 2, « = 3; or, x = ,ij = — , z = g
22. a; = 1, y = 2, z = 4 ; a; = — 1, y = — 2, s = — 4 ; x — 9,
y = — 6, s = 4 ; or, x = — 9, y = 6, z = — 4.
Art. 322; pages 246 to 248.
4. 12 and 7, or — 12 and — 7. 5. 11 and 7, or — 11 and — 7.
6. A, $2025; B, $900; or, A, $900; B, $2025.
ANSWERS TO EXAMPLES. 445
7. A, 25 ; B, 30. 8. Length, 150 yds. ; breadth, 100 yds.
9. 13 and 6. 10. A, $15; B, $80. 11. 10 lbs., at 8 cts.
12. A, $ 5 ; B, $ 120. 13. Duck, $ 0.75 ; turkey, $ 1.25.
14. Price, $ 1600 ; length, 1G0 rods ; breadth, 40 rods.
15. Larger, 864 sq. in. ; smaller, 384. 16. A, $ 275 ; B, $ 225.
17. 1st rate, 7 p.c; 2d, 6. 18. A, 40 acres at $ 8 ; B, 64, at $ 5.
19. Distance of towns, 450 miles ; A, 30 miles a day ; B, 25.
20. 3 and 1 ; or, 2 + sj 7 and 2  y/ 7. 21. Larger, 12 ft. ;
smaller, 9. 22. Width of street, 63 ft.; length of ladder, 45.
23. B, 15 days ; C, 18 days.
24. Length, 16 yds. ; width, 2 yds.
Art. 328 ; page 253.
3. (a: + 60) (a; + 13). 6. (x + 13) (a? 3). 9. (4a;l) (2a; + 5).
4. (x 9) (x 2). 7. (jc5)(2a; + 3). 10. (x  3) (4 x  3).
5. (a; 10) (a; + 6). 8. (7 a; + 3) (3 a; + 7). 11. (a; + 2) (2 a; 3).
12. (3a:2 + v/3)(3a^2 v /3). 13. (y/17 + 4 + a) (^1743:).
14. (7a3 + l + 2 v / 5)(7a; + l2v/5).
Art. 329 ; page 254.
2. ar + a; = 2. 5. 3 x 2  2 x = 133. 8. 3 x 1 + 17 x = 0.
3. a; 2 9 a; =20. 6. 21.x 2 + 44 x = 32. 9. a: 2 2a; = 4.
4. 5a; 2  12a; =
9. 7. 6ar+35a;=49. 10.
Art. 330 ; page 255.
x 2 — 2 m x=n— m 4
7. 0or y .
8. or — 4. 9. or ± 3.
in 5 x
lL*orl
a c
12. ± 2 or ± 3.
1 5
13 3° r± 2
14. ±sja
or — . 15. 0, 
5 7 1
■2'3'° r ~4
16. 2,3,3, 4, I, or 5.
446 ALGEBRA.
Art. 331; page 256.
3. (x + y/2 x + 1) (x  \/2 x + 1).
4. (a; + \Jx + 1) (x — \J x + 1).
5. (a + ^5lTb + b) (a  \j~h~ab + b).
6. (x°~ + 3 x y + f) (x 2 3xy + f).
7. (x + 1 + S/Sx + 2) (x + 1  \J~3x~+2).
8. (m 2 + mn + w 2 ) (m 2 — m n + to 2 ).
Art. 332 ; page 256.
2.
7. ^ 7 * V 71  1 ™ v / 7±V /: l
or
2^2 2^2
Art. 357; pages 269 and 270.
1. 4. 2. 11. 3. £. 4. 1 J. 5. ±4. 6. ± 12.
7. ± 14. 8. 25 and 20. 9. 23 and 27. 10. 12 and 15.
11. 8 and 18. 12. 26 and 14. 13. 17 and 12. 14. 12 and 8.
15. First, 1:2; second, 2 : 1. 16. Females : males = 4:5.
17. 8 : 7.
Art. 365; page 273.
2.4. 3. „ = 8* 4* 5.4. 6. y= 14 . .
°> 4 — 5cc
7. 10 inches. 8. 3 (y/ 2 — 1) inches. 9.143.
ANSWERS TO EXAMPLES. 447
Art. 370; page 276.
3. 1=71, £=540. 4. Z=69, £=620. 5. 1=57, £=552.
OQ A9 3
6. l=U5, £=2175. 7. Z=^,£=y. 8. l=,S=0.
9. l=^,S=±. 10. l=~,S=~. U.l=5,S=17.
4 2
Art. 371; pages 278 and 279.
q 95 1
4. a = 3,£ = 741. 5. a = ^,i==^. 6. <Z=,£=39.
7. d = — i, « = ?. 8. a=o,d=3. 9. w=18, £=411.
12 4
10. «Z = — 8, n = ll. 11. w = 30, Z = 80. 12. ^ = 52, a = 4;
or, » = 43, a — — 5. 13. n = l%l =  43.
Art. 372; page 279.
7 8 10 11 5 3 1 1
A 3'3"I'"3"" J, 2' w, 2' '2' ' 2'
4 2 3 _ 4 _ 5 5 Jl ^ _^ _!? ^ ??
*■ ^ — ^ 4 > ° 7 ' 7' 7' 7' 7' 7
2 6 14 22 „ am + 5 a(m l) + 2b
5'5' 5 ' 5 ' ' m + 1 ' m + 1
Art. 373; page 281.
3. 2500. 4. Last payment, $ 103 ; amount, $ 2704.
5. 4. 6. After 9 days, at a distance of 90 leagues.
7. 4, 11, 18, and 25. 8. 3. 9. 0. 10. 20 miles.
11. 2, 6, 10, and 14 ; or,  2,  6,  10, and  14. 12. 8.
448 ALGEBRA.
Art. 378; page 284.
4. 1 = 2048, £=4095. 9. 1 =  — , s = ^^
64 ' 192
64 2059 _ 1 511
5 l = m> S= 243. 10. 1 = ^, S=~.
6. Z = 2048, £=1638. 11. l^~, £=^.
7 l L s 3 ^ 12 Z 1 <? 341
*~ 256' ^256' "' ^~768'^~256
1 2047
8  l = zm> S =mi 13^ = 192,5=129.
Art. 379; page 286.
 1 c 341 2 _ 2
4. a = ^, £ = 7 r. 5. a = 7., Z =
2' 2 "" " _ 3' 6561'
6. r = 3, £=2186; or, r = 3, £=1094.
1 2457
7. ^j, /S= m . 8. » = 5, £=121.
9. n = l, r = \. 10. w = 6, Z = ?S 1L w = 8, a = l.
^ 2
Art. 380 ; pages 287 and 288.
3. 4.
5 3
160
y  19"
4 ?
3
ft 15
Art. 381
8 'i'
; page 288.
,0.12.
3.. 4.
13
27"
5 n
5  15
ft 8G 7
6  165* 7 '
17 237
150* 1100
ANSWERS TO EXAMPLES. 449
Art. 382; pages 289 and 290.
48 16 32 64 392781243 39
'3' 9' 27' 81' 243" 2' 2' 2 ' 2 ' 2 ' ° r ' 2' 2'
, , y. 5.  6, 18,  54,  162,  486,  1458.
927 81243 33333 3 3
~4' 16' ""64' 256* 4' 8' 16' 32' 64' 128' 256'
3 3 _3_ _3^ 3_ _3_ _3_
° r '~4' 8' ~16' 32' 64' 128' " 256'
Art. 383; page 291.
3. $ 64. 4. $ 295.23. 5. 3100 ft. 6. 5, 10, 20, and 40 ;
or,  15, 30,  60, and 120. 7.  4. 8. T V
Art. 386 ; page 292.
Z .E. 3. i. 4. 3 " •»
31' 78' ' 4' ' arc&?i + 2&a'
2.
Art. 387; page 293.
48 24 16 12 48 8 48
125' 65' 45' 35' 145' 25' 155'
5_5_5 ■ 21 _7 21 21
3 ' 4'" 3'" 2" '" '" 5 ' 3' 13' 17"
_ (m + 1) a 6 (m + 1) a b (m + 1) ab
m
b + a ' m6 + 2a6' mH3«2i'
Art. 397; pages 297 and 298.
4. Of 4 letters, 360 ; of 3, 120 ; of 6, 720 ; in all, 1956.
5. 1680. 6. 3838380. 7. 358800. 8. 15120. 9. 120.
10. 35. 11. 15504. 12. 31824. 13. 77520. 14. 648.
450 ALGEBRA.
Art. 403 ; page 302.
5. 1 + 5 c + 10 c 2 + 10 c 3 + 5 c 4 + c 5 .
6. a 6 + 6 a 5 a 8 + 15 a 4 x 6 + 20 a 3 x 9 + 15 a?x u +Qa x lh + x ls .
7. x s 8x 6 y + 24:X i 7fS2x 2 f+16y\
8. a 7 ^7a 6 i 6 c^ + 21a 5 i 5 cV 2 35aH 4 c 8 ^ 3 +35aH 8 c 4 ^ 4
21a 2 & 2 c 5 d 5 +7a&c 6 a ,6 c 7 cf.
9. m 12 + 18 m 10 w 2 + 135 m 8 w 4 + 540 m 6 w 6 + 1215 m 4 w 8
+ 1458 m n 10 + 729 w 12 .
10. a  10 20 a 8 a^+160 ar 6 x 640a~ ^+1280 or 2 ar1024a^.
11. c™ + 8 c*' cfl + 28 c 4 e^ + 56 c$~ <fi + 70 c% d 3 + 56 c 2 d^'
+ 28 c* <# + 8 J eft + d 6 .
12. m~^ + 14»"^ w 3 + 84 m 8 » 6 +280 m  ^" w 9 + 560m~*» 12
+ 672 m"^ w 15 + 448 j»~* w" 4 128 w 21 .
13. a 4 4ffl 3 i 2 x^ + 6 a~ 2 Z> 4 a;^  4 or 1 6 6 a; + b s sA
Art. 404; page 303.
2. 5005 a 6 a; 9 . 4.  19448 c 10 d 7 . 6. 42240 x~ 3 yK
3.2002 m 6 . 5.495 a 8 . 7. 262440 a 2 ar 7 .
Art. 405 ; page 304.
2. 1  4 x + 2 a; 2 + 8 x 3  5 x i  8 x h + 2 x 6 + 4 x 1 + x\
3. a; 6 + 9 x 5 + 30 a; 4 + 45 x 3 + 30 x 2 + 9 x + 1.
4. l6a; + 6a; 2 +16a: 8 — 12a 4 24a; 8 8a: 6 .
5. l+5a; + 5a; 2 10a; 3 15a; 4 +lla; 5 +15x 6 10a; 7 5x 8 +5x 9 cc 10 .
Art. 414; page 309.
3. l2a: + 2a; 2 2a: 3 + 2a; 4
4. 3 + 19 x + 95 x 1 + 475 x a + 2375 a; 4
5. 2a; + 3a: 2 a: 8 + 3a: 4
ANSWERS TO EXAMPLES. 451
6. l2a; + 2a; 3 — 2x 4 + 2x 6
7. l2x + ox 2 16x 3 + 4:7x i
1 5x 7_x* lTjc 3 31_x 4
9. 2 7 a; + 28a; 2  91 a; 3 + 322 a; 4
2 a; 7 a; 2 13 a; 3 8 a; 4
io. i + — — 27 + 81
1 3jc a 2 15k 8 49a 4
1L 2 + ~T + IT* 16 + 32
Art. 415; page 310.
2 a; 2 4 a; 1 _8_ 16 a; 32 a; 2
2  _ 3 _+_ 9 _+ 27 + 81 + 243
3, £C  1 + 3 + 2x5a; 2 16a; 3
4. x~ 2 — x 1 2 x + 2 a; 2 4 a; 3
Art. 416; page 311.
a; a; 2 a; 3 5 a; 4 a; 3 a; 2 3 X s 3 a; 4
3 * 1 + 2~8~ + 16~128"" ,1+ 2 + 8 "16 + 128"'
a; 2 a; 3 5 a; 4 _ a _ s 2 5 a; 3 10 a; 4
3. l_aj_— —  — g... «>. i 3 9 ~ 81  243 •'•
x 4 „ . a; 2 a: 2 13 a; 3 8 a; 4
4. l_a; + a; 2 + a; 3 + y ... 7. 1 + +— — ^ +^3 ...
Art. 418 ; page 314.
3 2.41 _J_ 6_
2  x~T2 + x~=2' x2 x 'x7 x6'
, 3 2 ,1 J_ 7 _2 3_
«~T+8" & '^T4 + a; + l' '"2»5 Sas + l"
Q 1 2 _J 1 1 4
8 "3+T£ + 3=!T y "6(x + l) 2(a;l) i "3(a;2)'
452 ALGEBRA.
Art. 419 ; page 316.
o 11 1 „ 1 4 4
2 — ^ + 7— ^Tv 2 + 7— "TV, 4.— +  — +
x + l (x + l) 2 (x + l) 3 ' x2 (x2f ' (a; 2)
3 2 3 5 3 6
B'
x5 (x5) 2 ' x + l (x + l)' 2 (x + l) 3 '
3 5
6.
7.
2 (2 a;  5) 2 (2 x  5) 2 '
2 4 3
3 a; + 2 (3x + 2) 2 (3x + 2) 3
Art. 420 ; page 317.
_ 2 3 5 .515
2. — , — ; . 4.
x x+2 (x+2) 2 ' 'x x* x + l (x + l) 2 '
„11 1 1 .123 4
3. +— T + —0 + 7— 2K3 5. ,+
x x— 1 x—2 (x—2)' 2 ' ' x x 2 x s x + 5*
■i
x2 2x3 (2x3)
„ 5 1 2 5 4
7. S + T—
X X 2 X 3 X + l (x+l) 2 '
Art. 422; page 320.
o 2,84 /i y 3?/ 19?/ 6 19 y'
3. x = y — ?/ 2 + ?/ 8 — y 4 . . . 4. x = sL j £ — . .
J J J J 2 16 + 128 128
5. x = y + y 3 + 2 y 5 + 5 y 7 . . .
vy ; 2 i 3 4 ■"
7 x _ ?/ , ^,2^ 17 y' y 2y 2 y 3 14y*
^~ y+ 3 + l5 + 315" 8 ' *3 + 27~~243~2T87"
Art. 425; page 325.
 § 5  .15 1,5 a . 5 _a
4. a 2 + a 2 x + ^a 2 x' + z—a 2 x z — — a ^ 4
«s 8 lb 128
ANSWERS TO EXAMPLES. 453
5. 1  6 x + 21 x 2  56 x 3 + 126 x*
3 12 2 52 3 234 4
b. i + 5 x + 25 x + 125 x + G25 x
i 1 _i. 1 _ 1 _i 5 z
7. a J —^a *x — ^a 'X' — zr^a 2 x 6 — zr^a J x*
£ o lb lwo
„ , 1 2 „ 14, 35
8.1 x +  x * x , + mX >
9. ar 3 + 3a 4 x + 6 a~ 5 x 2 + 10 a~ 6 x 3 + 15 «" 7 x* ....
10. c"^— c 3 df + c~* d 2  c 6 d 3 + c" 1 *' d*
11. x z —2x b y — X s y 2 — X s y 3 — ^x 3 y*
o o
I 3 15 7 35 3 a 315 jjl 6
14. m + 6 in 6 n + jr m J n 6 + = mr n 2 \ — — m J ?i . . .
2 2 o
in i i^ i ™ o o 1760 „ „ 12320
13. 1 — 10 xy 1 + 80 a; 2 ?/ 2 qb'jt'H 5— a? 4 ?/" 4
2 2 8
15. a 4 + 12 a 5 gT 2 + 90 a 6 y~ 4 + 540 a 7 jr 6 + 2835 a s y
— 8
Art. 426 ; page 326.
33 aT'^'x'' 315 a 8 44 x^' y*
"~2048 ' 128"' 6561 '
663 x "V
4. 84 m 6 . 6.  " y ■ 8. 210 w^c" 8 .
8192
9. _?5? a ^aj 6 . 10. 3§x 3 «y Vi z~ 1 £.
Art. 427; page 327.
3. 3.14138. 5. 9.94987. 7. 2.03054.
4. 2.08008. 6. 1.96101. 8. 2.97183.
454
ALGEBRA.
2.
3.
1 + x
Art. 435; pages 331 and 332.
4 11 x „ 2 + 5« + 5a 2
1 — x — x 2 '
a
b + ex
4.
5.
1 — 5 x + 6 x 2
1+x
1  2 x + x 2 '
6.
7.
(1 + a) 3 '
3 — x — 6 x 2
l2a;a: 2 +2a; 3
3. 3.
8. 225.
8.
l + 2cc
1 — x — x 2
9.
2 + 2 a;  3 x 2
l — x + x 2 — x &
4.
Art. 440; page 336.
14. 5. 30. 6. 1365.
7. 5050.
9.
n i + 2n z + n 2
11. 165.
10.
6 n 6 + 15 m 4 + 10 w 8
?«
30
12. 5525.
3. 4.0514.
1. 1.681241.
2. 2.644438.
3. 1.748188.
Art. 443 ; pages 338 and 339.
4. 3.634241. 5. 2.23830. 6. 44.24.
7. $1,356.
Art. 455 ; page 344.
4. 1.991226. 7. 2.225309.
5. 1.924279. 8. 3.848558.
6. 2.753582. 9. 2.702430.
10. 3.489536.
11. 4.191785.
12. 4158543.
1. 1.176091.
2. 2.096910.
3. 0.154902.
Art. 456 ; page 345.
4. 2.243038. 7. 0.853872.
5. 0.522879. 8. 1.066947.
6. 1.045758. 9. 0.735954.
Art 464; page 350.
2. 8.72427610. 4. 9.47070410. 6. 1.527511.
3. 1.714330. 5. 0.011739. 7. 8 78021010.
ANSWERS TO EXAMPLES.
4jj
8. 4.812917. 11. 9.94255010. 14. 4.89381.
9. 7.01315010. 12. 3 863506. 15. 1.718451.
10. 2.960116. 13. 8 64040910. 16. 7.498424010.
17. 9.27537410. 18. 1.9792784.
2. 76.
3. .2954.
4. 6.61005.
5. 55606.5.
6. .011089.
Art. 465 ; page 352.
7. 186 334. 12. .034277.
8. .223905.
9. 1000.06.
10. 9.77667.
• 11. .00130514.
17. .00548803.
18.
13. 46.7929.
14. 11.327.
15. 8.63076.
16. .2070207.
734.9114.
Art. 466; pages 353 and 354.
1. 2.125240. 4. 3 108462. 7. 9.613158  10.
2. 8.22396210. 5. 9.59416110. 8. 9.97003610.
3. 9.85216910. 6. 7.31532110. 9. 9.90523210.
Art. 468 ; pages 356
1 to 358.
1.
.0341657.
13. 1.70869.
25. .580799.
2.
.650573.
14. .788547.
26. .631188.
3.
13560.2.
15. .680192.
27. 83.5656.
4.
.136085.
16. 2.24328.
28. .297812.
5.
1.14720.
• 17. .296850.
29. 98.4295.
6.
1.41421.
18. .191680.
30. 1.65900.
7.
1.49535.
19. .644849.
31. 3 07616.
8.
.0655264.
20. .501126.
32. .867674.
9.
1.97221.
21. 1.09872.
33. 2.09389
10.
458.623.
22. 1.06178.
34. 46809.2.
11.
.000113607. 23. 1.09328.
35. .588142.
12.
5.88336.
24. 1.65601.
36. 1.80446.
37.
.00323011.
38. .0334343.
456
ALGEBRA.
The following are the values of the expressions in Art. 468,
when calculated by sevenfigure logarithms :
1.
.034165G8.
13.
1.708689.
25.
.58079S7.
2.
.0505727.
14.
.7885469.
26.
 .6311888
3.
13560 27.
15.
.6801947.
27.
83.56558.
4.
.1360851.
16.
2 243284.
28.
.2978123.
5.
1.147203.
17.
.2968501.
29.
98.42J91.
6.
1.414214.
18.
 .1916795.
30.
1.658989.
7.
1 495349.
19.
.6443490.
31.
3.076162.
8.
.06552632.
20.
.5011282.
32.
.8676754.
9.
 1.972211.
21.
1.098718.
33.
 2.093891.
10.
458.5759.
22.
1.061780.
34.
46808.95.
11.
 .0001136063
. 23.
1.093280.
35.
.5881412.
12.
5 883366.
24.
1.656005.
36.
1.804459.
37. .003230121
38.
,03343431.
Art. 469 ; page 359
3. ,
458156.
5.
 .494903.
7. 
2.70951.
4. .
185339.
6.
 .260231.
8. 
 10.2341.
The results with sevenfigure logarithms are as follows •
3. .4581568. 5.  .4949028. 7.  2.709513.
4. .1853394. 6. .2602272. 8. 10.23414.
Art. 479 ; pages 368 and 369.
1. 7.
3.  6. 5.
7.
7. 6.
2. 6.
4 .4 e.
5.
8. 7.
9. 1.56937.
13. 11.725 yrs.
17.
3.96913.
10. 2.44958.
14. $9756.59.
18.
7.18923.
11. 2.00906.
15. 7 per cent.
19.
 2.4578.
12. $5421.33.
16. 9.392 yrs.
20.
 1.07009
ANSWERS TO EXAMPLES. 457
The results of the last 12 examples, using sevenfigure
logarithms, are as follows :
9. 1.569369. 13. 11.725 yrs. 17. 3.969124.
10. 2.449576. 14. $9756.59. 18. 7.18922.
11. 2.009056. 15. 7 per cent, 19. 2.457802.
12. $5421.35. 16. 9.392 yrs. 20. 1.070092.
Art. 489; page 373.
2. 3 and  5. 3. a and  ( 1 ± y/^3). 4. 2 and 2. 5. ± 4.
6 . X 3_ 6a .2_ 6a ,_ 3 = 7> ? and 5 8 . ?andl
3 2 4 5
Art. 490; page 374.
2. z 3 + 9x 2 + 23£ + 15 = 0. 4. 6ai 3 lla: 2 + 6*l = 0.
3. a 8 19 a: 30 = 0. 5. cc 4  5x' + 4 = 0.
6. a; 4  10 x* + 35 x 2  50 cc + 24 = 0.
7. cc 3  13 x 2 + 56 x  80 = 0.
8. x 4 6a; 3 +5x 2 + 12x = 0.
9. 12 z 4 + 55 r 3  68 x 2  185 x + 150 = 0.
Art. 494; page 375.
5
1. Sum, ; product,  6. 2. Sum,  ; product, 12.
3. 2±2v/2.
Art. 504; page 382.
2. ?/ 3 + 24?/ 2 + 191?/ + 498 = 0. 3. y*  6?/  if + 55y  76 = 0.
P
■:
Art. 505; page 383.
2. y 2 ^ + <7 = 0. 4. ^15y+26 = 0.
y 3 + "2y ==0, 5 * 2/ 4 6y 2 137/9 = 0.
458 ALGEBRA.
Art. 513; page 388.
2. 1, 1, and 6. 4.1,1,1, and 3.
3. 2, 2, and 3. 5. 2, 2, 2, and  6.
Art. 517; page 390.
2.  1, 1, and 5. 3. 3. 4. 1. 5. 2.
Art. 520; page 392.
3. 1 + s/U. 4. 1 + V 15  5  (1 + V^) 6.  (1 + $5).
Art. 527; page 399.
3. Three ; respectively between and 1, 1 and 2, and — 1 and — 2.
4. Three ; two between 1 and 2, and one between — 3 and — 4.
5. One ; between 2 and 3.
6. Four ; respectively between and 1, 1 and 2, 2 and 3, and
— 2 and — 3.
7. None.
8. Two ; respectively between 2 and 3, and 3 and 4.
Art. 532; page 403.
3. — 1,  2, and  3. 9. A, and 1 ± }f—i.
4. 2, 2, and 3.
5. 2, 4, and  1 ± y/^3.
6.  , 4, and — ^ •
7. 2, and — — * .
8. 3. 6, and  2.
10.
1,
2, and 3.
11.
3
9
, and ± 2
v/
2.
12.
o
13.
3.
14.
3,
4. — 3, and ■
— 5.
ANSWERS TO EXAMPLES. 459
Art. 538; pages 407 and 408.
2 1 9 ±x /77 3 ± y/5 l_^ ±V /^_ 2j? 3
3. 1,1,1, or =^?. 6. 2,i,3,or.
4. ±l J±V /=T,or^f: 3 . 7. l,5,i or2±^3.
y/33 5 ±^42 10 y/33  y/33  5 ± y/42 + 10 y/33
8  4 ' ° r ~I~
q _i 1 + y/ 5 ±V /2 V /5jri ~ Q " or 1^5+^2^510
a> *' 4 4
l^5±y^51 l +v /5±y/2v/510
iu. w, 2~ ? 01 o
Art. 541; page 410.
3. 3or^±£3. 7. lorl±YE5.
2 2_
4. 4 or 1 ± 4 \/=3. 8. 3 or 1 ± ^~ 3 '
5. 3, 3, or 2. 9. 2, 2, or— 1.
6. 1,1, or 11. 10. $±1(2.
Art. 550; page 417.
2.2.09455. 3.7.61728. 4. 1.3569, 1.6920, and 3.0489.
5. 14.95407. 6. 2.2674 and 36796.
7. 2.85808,60602, .44328, and  3.90738.
Art, 551; page 419.
2. 3.864854. 4. 2.4257. 6. 10.2609.
3. 4.11799. 5. .66437. 7. 8.414455.
Art. 552; page 420.
2. 153209. 3. 1.02804.
T A B L E
CONTAINING THE
LOGARITHMS OF NUMBERS
FROM 1 TO 10,000.
No.
Log.
No.
Log.
No.
Lo£.
No. Log.
No.
Loft.
1
0.000000
21
1.322219
41
1.612784
61
1.785330
81
1.908485
2
0.30103C
22
1.342423
42
1.623249
62
1.792392
82
1.913814
3
0.477121
23
1.361728
43
1.633468
63
1.799341
83
1.919078
4
0. CO 20 GO
24
1.380211
44
1.643453
64
1.806180
84
1.924279
6
0.698970
25
1.397940
45
1.653213
65
1.812913
85
1.929419
6
0.778151
26
1.414973
46
1.662758
66
1.819544
86
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OF NUMBERS.
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3
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58
4
1573
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58
6
2156
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2506
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1946
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4
1458
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1610
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51
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2118
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2322
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1362
1412
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1660
1710
1760
1809
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OF NUMBERS
15
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2
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1435
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1532
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1726
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5
1823
1872
1920
1969
2017
2066
2114
2163
2211
2260
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2308
2356
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2502
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2647
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1421
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1611
1658
1706
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1801
1848
47
6
1895
1943
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2038
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2132
2180
2227
2275
2322
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2369
2417
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2559
2606
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1276
1322
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1415
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1740
1786
1832
1879
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1971
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9 9565
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9783
9826
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9957 43
nTT
1 o
1
2
4
J &
6
7
8
9 ID.
14 DAY USE
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